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For a school project I am making a gravitational computer model. As I want to Milankovitch cycles, I need to calculate the eccentricity of an orbit after the model has completed its simulation.

I have thought about calculating the eccentricity using the aphelion and parohelion height, but these are not available as it is a simulation and as it therefore stores the data for one point on the ellipse.

The data I have after the model has completed:
X,Y,Z coordinates of points that it has saved
X,Y,Z velocities of points that it has saved

I have been able to calculate the semi-major axis using the vis-viva equation. I have tried to get the value using the vis-viva equation but I got stuck as I think I need at least the Aphelion or Parohelion to calculate the eccentricity.

My question: Does anyone know how to calculate the eccentricity at any given point in the orbit using the velocity and coordinates (without simulating all points around it)?

If not, does anyone maybe have a better idea/suggestion that I can try to gain the eccentricity values?

If you have the relative position $$mathbf{r}$$ and velocity $$mathbf{v}$$ at the same point in time, compute the eccentricity vector:

$$mathbf{e} = frac{mathbf{v} imes mathbf{h}}{mu} - frac{mathbf{r}}{|mathbf{r}|}$$

Where $$mathbf{h} = mathbf{r} imes mathbf{v}$$ is the specific relative angular momentum and $$mu = GM$$ is the standard gravitational parameter.

The eccentricity is then given by the norm of the eccentricity vector $$left| mathbf{e} ight|$$.

One uses the so called eccentricity vector, also called (up to a factor) Laplace-Runge-Lenz vector.

Given the position $$vec{r} = egin{bmatrix}x y zend{bmatrix}$$ of the object at given time $$t$$ and its velocity $$vec{v} = egin{bmatrix}u v wend{bmatrix}$$ both with respect to an inertial coordinate system cantered at the origin of the gravitational field, the eccentricity of the orbit (Keplerian case) can be calculated by first computing the eccentricity vector:

Step 1: Calculate the angular momentum $$vec{L} = vec{r} imes vec{v}$$ Step 2: Calculate the eccentricity vector $$vec{e} = frac{1}{mu}, ig(,vec{v} imes vec{L},ig) - frac{vec{r}}{|r|}$$ Step 3: Calculate eccentricity $$e = |vec{e}| = sqrt{vec{e} circ vec{e},}$$ where $$cdot circ cdot$$ is the dot product, e.g. $$|r| = sqrt{vec{r} circ vec{r},} = sqrt{x^2 + y^2 + z^2}$$

## Calculating the eccentricity of an exoplanet

I'm wondering how to calculate the eccentricity of an exoplanet by its radial velocity vs. phase graph. To clarify my question I will take an exoplanet called WASP-14b 2 as an example (http://exoplanets.org/detail/WASP-14_b).

A plot of the radial velocity of the star vs the phase is displayed in the upper left corner. I am wondering how I could possibly calculate the eccentricity of the exoplanet using this graph (or some other values given in the original measurements). I found a few ways to calculate the eccentricity:

This uses the eccentricity vector which is calculated using this formula:

The problem here is that this formula needs the specific angular momentum vector and the position vector, which I do not know given only the measurements. However, there is another way to calculate the eccentricity:

where $r_a$ is the radius of the apoapsis and $r_p$ the radius of the periaosis. These values are not known using only the measurements, but I believe it should be possible to calculate them by taking the integral of the sine function (radial velocity vs. phase). This would give me the position of the star at any given moment. The problem is that I cannot find the exact points displayed in the graph anywhere, let alone a sine function that would fit them.

When I do get an integral of the function I still have to create one for the planet itself, since this describes the movement of the star. I am able to calculate the mass of the planet using the following formula:

which gives me the distance between the star an the planet. Next I can calculate the velocity of the planet using:

And after that I can calculate the mass of the planet using this formula:

But this is where another problem comes up as a Wikipedia article on Doppler Spectroscopy states: "Observations of a real star would produce a similar graph, although eccentricity in the orbit will distort the curve and complicate the calculations below."

Where do I find the corrected calculations and how can I possibly calculate the eccentricity of this planet using only these values ($M_$ and the plot, of which I cannot find the exact points)?

## Calculation of Eccentricity of orbit from velocity and radius - Astronomy

The Radius of Perigee calculator computes is the distance between an object (often a planet) and orbiting body when the orbiting body is at perigee based on the semi-major axis (a) and the eccentricity (e) of the orbit.

INSTRUCTIONS: Choose units and enter the following:

• (a) This is the semi-major axis of the orbital ellipse .
• (e) This is the eccentricity of the orbit ( ellipse )

Radius of Perigee (rp): The calculator returns the radius of perigee in kilometers. However this can be automatically converted to other distance units including a full suite of astronomical distance units (e.g. light years, parsecs, astronomical units) diagram illustrating various orbital characteristics,

Related Calculators:

#### The Math / Science

Perigee is the point in an orbit at which the orbiting body is closest to the object it orbits. Thus, the Radius of Perigee represents the minimum distance between an orbiting body and the object it orbits. Radius of Perigee may be contrasted to Radius of Apogee , which is the maximum distance between orbiting and orbited bodies.

This equation produces the Radius of Perigee rp, based on length of semi-major axis (a) and eccentricity (e) of orbit. Distances are measured from the centers of bodies.

Example Parameters

Orbit Semi major axis Eccentricity
Earth (about Sun) 1 149.60 x 10 6 km 0.0167
Moon (about Earth) 2 0.3844 x 10 6 km 0.0549
Saturn (about Sun) 3 1,433.53 x 10 6 km 0.0565

#### Remembering the Difference between Perigee and Apogee

Since apogee and perigee are not commonly used terms, they are often confused. A simple memory convention to remember the difference is this, count the number of letters. Apo and far, have three letters. Peri and near have four letters. Therefore perigee is the nearest point in the orbit, and apogee is the farthest point in the orbit.

### History

The understanding of orbital perigee arose directly from Kepler&aposs laws of planetary motion published in 1609 4 .

That was a whole lot more work than was needed to answer those questions!

Specific orbital energy is indeed ##-frac <2a>## , but it's also ##frac 1 2 v^2 - frac r## . You don't need to find the semi major axis to find the specific orbital energy. Just compute it directly. What's the sign of this result? If it's positive, the comet is on an escape trajectory. If it's negative, it's in a bound orbit.

Regarding the nature of the orbit, what's the circular velocity at 13.75 au? If the comet's speed is something different than this, the orbit can't be circular.

So I used E=K+U = 0.5v^2 -GM/r to find the orbital energy and I got E=-6.46E7 (joules?)

By the way, do we just assume the mass of the asteroid is negligible in the energy eqn?

So then the asteroid is in a bound orbit since the energy is negative, right?

However, how would I find the circular velocity using just radius? I think the only thing I can compute for circular motion is a=v^2/r.

Ohhh would I use v_circular=sqrt(GM/r) ?

then i get v_circular=8035m/s which is much greater than v=61m/s
So that means its an ellipse?

The energy is negative, so it's in a bound orbit -- either circular or elliptical. You've correctly found that velocity is not that of circular orbit at that distance, so the orbit isn't circular. That leaves one choice.

Note that if you had been told the velocity was 8035 m/s the answer would have been "not enough information". All that this would tell you is that r=a. The velocity has to be equal to circular orbital velocity and normal to the radial displacement vector to have a circular orbit. You weren't told anything regarding the direction of the velocity vector.

Aside: Since you were not told anything about the direction of the velocity vector, you can't solve for the eccentricity. All you do is find a lower bound on the eccentricity (the upper bound is of course one). In this case, that lower bound is rather large. This is a highly eccentric orbit.

## Disclaimer: The following material is being kept online for archival purposes.

### (21a) Application of Kepler's 3rd Law

In these closely linked units, Newton's calculation is applied to artificial Earth satellites. It is shown that at least for circular orbits, this calculation leads to Kepler's 3rd law. The velocity required for a low Earth orbit is derived, and a practical formula is obtained for the orbital period in a circular Earth orbit of any radius.

Part of a high school course on astronomy, Newtonian mechanics and spaceflight
by David P. Stern

 This lesson plan supplements: "Kepler's 3rd Law," section #21 http://www.phy6.org/stargaze/Skepl3rd.htm "From Stargazers to Starships" home page: . stargaze/Sintro.htm Lesson plan home page and index: . stargaze/Lintro.htm

Goals : The student will learn

Terms: Orbital velocity, escape velocity, synchronous orbit.

Today we will apply Newton's calculation to circular Earth orbits. As already noted, all three of Kepler's laws can be derived from an inverse-square gravitational force, assuming only that it acts primarily between only two bodies--e.g. Sun and planet, Earth and Moon--while the pull of all others can be neglected.

With our limited math we can handle here only circular orbits, not elliptical ones. Circular orbits are a somewhat special and simple case. The first law for instance is automatically satisfied, because yes, the circle is an ellipse--of zero eccentricity and with both its foci combined at the center.

The second law also does not add any information about the motion. The speed of a satellite or planet in a circular orbit is constant, and so is its distance from the center. The area swept by the radius per unit time is then automatically constant.

The third law &, however, is meaningful even for circular orbits. Newton's calculation allows it to be checked, and today we will show that the inverse-squares law indeed gives the expected result.

In this day and age we are interested not just in planetary orbits but also in the orbits of artificial satellites. We will therefore use such orbits to illustrate the calculation, and some useful results will also be derived.

We start by calculating the velocity V 0 required by a spacecraft in a circular low Earth orbit--what is called in Russia "the first cosmic velocity." If g is the acceleration due to gravity and RE the radius of the Earth, we find
V 0 2 = g RE (Teacher may go over derivation)

Using the energy equation of the Kepler motion, one can also show that the " escape velocity" V 1 from distance RE ("second cosmic velocity" in Russia) is given by V 1 2 = 2g RE

It is larger by a factor equal to the square root of 2, which is 1.4142..
(This calculation may be included or omitted, depending on the level of the class and the time available). Questions and worked examples

This section is again mostly concerned with one calculation, and the questions asked are all related to it.

If you give the above velocity V 0 to a rocket launched vertically, straight up, how far from Earth will it get before falling back?
Hint: the semi-major axis of an orbit depends only on the energy.

[The rocket will rise and then come down along the same radial line, following what is essentially a very narrow and elongated ellipse (eccentricity close to 1).

Given the orbital velocity for a circular orbit at 1 R E as 7.9 km/sec, what is the escape velocity at the surface of the Earth (r = 1 R E )?
What sort of "escape" does it provide, and what sort doesn't it?

The escape velocity is V 1 = 7.9*SQRT(2) = 7.9*1.4142 = 11.2 km/s. A velocity greater than V 1 makes it possible for the launched object to escape Earth and not fall back. However, it will still be bound by the Sun's gravity, and will stay in an orbit around the Sun, similar to the Earth's.

If an object above the atmosphere has a velocity greater than the escape velocity V 1 in what upward direction must it move to escape the Earth's gravity?

It does not matter. An escape orbit is the limiting case of an orbit with very, very large semi-major axis, and therefore a very distant apogee. With a starting speed close to the escape velocity, in whichever upward direction the object is moving, its starting point is always close to the perigee, and it is always headed for the apogee. In the limit (=at escape velocity), this means it is heading for infinity.
The calculation is similar to the preceding one. Escape velocity from the Sun at 1 AU is
V 1 = 30*SQRT(2) = 30*1.4142 = 42.4 km/s.
Thus an additional velocity of (42.4 - 30) = 12.4 km/s is required.

(The question below is really just for debating fun!)

In some weird alternate universe (already met in the lesson on Newton's 2nd law) weight and mass are not proportional. Two materials, astrite and barite, have the same weight per unit volume, but a volume of astrite has twice the mass of a similar volume of barite.

Both are strong and light metals, and are a natural choice for spacecraft construction we can assume that barite behaves the way aluminum does on Earth. Which of the two would be a better choice?

Say an Earth-like planet in the alternate universe has a radius R E and surface gravitational acceleration g. In a circular orbit having a radius of R planetary radii, the weight of each satellite is the same, say
F = mg/[R/R E ] 2

The inertia of the astrite satellite is twice as large. So the centripetal force needed to hold it in orbit Requires a velocity U 1 satisfying F = 2m U 1 2 /R

while the barite satellite, with smaller inertia, needs U 2 satisfying F = m U 2 2 /R

--The planet Mars has a radius of R M =3390 km and its satellite Deimos orbits it in a near-circular orbit with orbital period T D = 1.26244 days and a mean distance of R D = 23,436 kilometers. What is (1) The acceleration g M due to gravity at the surface of Mars and (2) The escape velocity there?

If the acceleration on the surface is g M , then for a circular orbit at distance R D

where v is the velocity of Deimos in its orbit. From a previous calculation, if N is the number of orbits per second and T D is also given in seconds

T D = 1.26244*86400 sec = 109,075 sec.

v = 6.2832*(23,436,000 m)/109,075sec = 1350 m/sec

In calculating a ratio we can use kilometers, so R D /R M = 23,436/3390 = 6.9133

g M = ((1350) 2 /23,436,000)*(6.9133) 2 =

i.e. slightly more than 1/3 the acceleration of free fall on Earth. The orbital velocity V 0m at the surface of Mars, in analogy with V 0 derived for Earth, is found from

V 0m 2 = g M R M = 3.717*3,390,000 = 12,600,000

V 0m = 3549.6 m/s = 3.5496 km/s

According to Kepler's 3rd law, T is proportional to a , where T is the orbital period in seconds and a the semi-major axis in meters. That implies T 2 = ka 3 . In a circular orbit around Earth, a=r where r is the orbital radius. Can you derive k for such orbits?

The result is derived in "Stargazers" and is k = g R E /4π 2

Here the teacher may continue with lesson 21a (which is optional). Some questions related to that lesson (in all that follows, * marks multiplication):

From Kepler's 3rd law, the orbital period T around Earth in a circular orbit at distance R is

--At what distance R is the orbital period around Earth 24 hours? Why is that orbit important?

We have 86400 sec = 5063.5 R*SQRT(R) = 5063.5 R 3/2

R 3/2 = 17.0633 R = (17.0633) 2/3 = 6.628 R E

It has been speculated that if we could manufacture a lightweight cable of near-infinite strength, such a satellite could be anchored to the spot below it on Earth, and loads could be passed to it by an elevator guided by that cable, making leaving Earth essentially independent of rockets.

As payloads were raised, they would need a motor to raise them along the cable, but the energy can be applied gradually. As payloads rise, the rotation of Earth would also be very, very slightly slowed down.

## Calculation of Eccentricity of orbit from velocity and radius - Astronomy

Problem #1: The Kuiper Belt is a collection of comet-sized junk which orbits the Sun with nearly circular orbits of typical radius 35 A.U. It is surmised that this reservoir of material is the source of the the so-called short period comets which occasionally transit the inner solar system. What is the period of the nearly circular orbits of these objects?

Solution: Oh no! It's the return of Kepler's Third Law! As you no doubt recall from the beginning of this course several eons ago, Kepler's Third Law is a relation between the period and semi-major axis of objects that orbit the Sun:

Hopefully, you also remember that in order to use this particularly simple expression of Kepler's Third Law, you need to express periods in years (Earth years, that is), and semi-major axes in A.U.

The problem tells us the radius of the nearly circular orbits of the Kuiper Belt objects is about 35 A.U., but this is a radius, not a semi-major axis, so what are we to do? Have no fear just recall that a circle is a special case of an ellipse with eccentricity zero --- that is, the major and minor axes are the same size. Thus, for a circle the radius, major axis, and minor axis are all the same quantity, and therefore we can use 35 A.U. as the semi-major axis in this problem. So,

and so the period is the square root of this quantity,

Problem #2: In order for a Kuiper belt object to become a short period comet that enters the inner solar system, the eccentricity of its orbit has to change from about 0 to pretty close to 1. Suppose this magically happens, and the orbit of a Kuiper belt object changes from a nearly circular orbit with radius 35 A.U. to a highly elliptical orbit with eccentricity of nearly 1. Suppose further that the new orbit has aphelion (point farthest from the Sun) of 35 A.U., and perihelion (point closest to the Sun) of just 0.8 A.U.

a) What is the period of this new comet?

b) What is the eccentricity of his comet's orbit?

In the above picture, I've drawn the Kuiper Belt objects as little blue circles orbiting the yellow Sun with circular orbits (i.e., eccentricity = 0) at a distance of a 35 A.U. I've also drawn in the orbit of a highly eccentric comet (eccentricity = almost 1), whose perihelion is 0.8 A.U., and whose aphelion is 35 A.U.

Are these reasonable orbits for objects around the Sun? Recall Kepler's First Law, which states that objects in orbit around the Sun orbit in ellipses with the Sun at one focus. The circular orbits of the unperturbed Kuiper Belt objects satisfy this requirement, since the Sun is at the center of the circles, and since a circle (a.k.a. ellipse of eccentricity zero) has both foci located together at its center. Now what about this cometary orbit? We're told that the orbit has an eccentricity of nearly one, which means that f/a is nearly one, where f is half the distance between the ellipse's two foci, and a is the semi-major axis. Well, if f/a is nearly one, then the foci must be located pretty far from one another along the major axis. In the drawing above, I've put the Sun way down at one end of the major axis, and since Kepler said we need to have the Sun at one of the foci of an object's elliptical orbit, it looks like this cometary orbit satisfies his first law. Note: Do not put the Sun at the center of an elliptical orbit. It absolutely screams out that you do not understand Kepler Laws.

Now compare the comet's orbit and the circular orbits of the Kuiper belt objects in the diagram above. You should be able to convince yourself that the major axis of the comet's orbit is equal to the aphelion distance (35 A.U.) plus the perihelion distance (0.8 A.U.), which is much smaller than the major axis of the circular orbits. We need the comet's semi -major axis, so that's half of this value. Once you're convinced yourself that the comet's semi-major axis is half of 35.8 A.U., or 17.9 A.U., the remainder of the problem is the same as problem #1:

and so the period is the square root of this quantity,

Sounds pretty close to Comet Halley's 76 year period, doesn't it?

Note that because the eccentricity is a ratio, it has no units.

Problem #3: Your textbook tells us that there are approximately 1 million asteroids of diameter 1 km or larger, and that together their mass is approximately 2% of the mass of the Moon. If we assume that the "average" asteroid is a sphere with diameter 10 km, then what is the average density of the "average" asteroid? Express your answer in km/m 3 so you can compare it to densities you know. Based on this result, what do you think most asteroids are made of?
Hint: Pretend that 1) all asteroids are exactly the same size and density, 2) that there are 1 million of them, and 3) together they account for the quoted mass.

Solution: This is an example of the kind of estimating astronomers do all of the time. In this case, we'll estimate the typical density of asteroids from knowledge of the mass of all of them, and a guess as to the size of a typical asteroid. We could instead measure the size and mass of each asteroid individually (though both measurements are really hard!), and then calculate a density for each one, but in most cases, we're more interested in an average value for the group than in each individual density.

For density, we need to know that mass and volume, so let's start by calculating the mass of the "average" asteroid. We're told that the mass of all of the asteroids together is about 2% of the mass of the Moon, which you can look up in your book:

• Mass of all asteroids = 0.02 x M Moon
• = 0.02 x (734.9 x 10 kg )
• = 0.02 x 0.0123 x 5.97 x 10 24 kg
• = 1.47 x 10 21 kg

But we're still not done with the mass calculation, though, since we want the mass of the "average" asteroid. The value above is the total mass of all the asteroids, but if that total is made up of one million "average" asteroids, then the mass of each average asteroid must be

• mass of "average" asteroid = mass of all asteroids / 1 million
• = 1.47 x 10 21 kg / 1 x 10 6
• = 1.47 x 10 15 kg

Now let's calculate the volume of an "average" asteroid. We're told that the diameter is 10 km, which is 10,000 m. We want the radius, which is half of the diameter, or 5,000 m. So the volume is:

Now, the density is simply

This density is higher than that of water (1000 kg/m 3 ) and less than that of iron (8000 kg/m 3 ), and is pretty similar to that of normal everyday rock. Thus, we can conclude that asteroids are probably made of regular old rock.

## Numerical Problems on Critical Velocity and Period of Satellite – 01

In this article, we shall study to solve problems to calculate time period and orbital speed of satellite.

Example – 01:

Calculate the speed and period of revolution of a satellite orbiting at a height of 700 km above the earth’s surface. Assume the orbit to be circular. Take the radius of the earth as 6400 km and g at the surface of the earth to be 9.8 m/s 2 .

Given: height of satellite above the surface of earth = h = 700 km, Radius of earth = R = 6400 km, G = 6.67 x 10 -11 N m 2 /kg 2 g = 9.8 m/s 2 .

To Find: speed of satellite = vc = ?, Period = T = ?

r = R + h = 6400 + 700 = 7100 km = 7.1 x 10 6 m,

The speed of a satellite orbiting around the earth is given by The time period of a satellite orbiting around the earth is given by Ans: The speed of the satellite is 7.519 km/s and the period of revolution of the satellite is 5930 s.

Example – 02:

From the following data, calculate the period of revolution of the moon around the earth: Radius of earth = 6400 km Distance of moon from the earth = 3.84 x 10 5 km g = 9.8 m/s 2 .

Given: radius of earth = R = 6400 km = 6.4 x 10 8 m, radius of orbit of moon = r = 3.84 x 10 5 km = 3.84 x 10 8 m, g = 9.8 m/s 2 .

To find: Period of revolution of moon = T =? Ans: Period of the moon around the earth is 27.3 days

Example – 03:

A satellite is revolving around the earth in a circular orbit in the equatorial plane at a height of 35850 km. Find its period of revolution. What is the possible use of such a satellite? In what direction is such a satellite projected and why must it be in the equatorial plane? Given g = 9.81 m/s 2 Radius of earth 6.37 x 10 6 m

Given: Radius of orbit of the earth =R = 6.37 x 10 6 m, height of satellite above the surface of the earth = h = 35850 km = 35850 x 10 3 m = 35.85 x 10 6 m, G = 6.67 10-11 N m 2 /kg 2 , g = 9.8 m/s 2 .

r = R + h = 6.37 x 10 6 m + 35.85 x 10 6 m = 42.22 x 10 6 m Ans: Period of the satellite is 24 hr.

Such a satellite is called geosynchronous satellite and is used for communication, broadcasting and weather forecasting. Such a satellite moves in the same direction as that of the rotation of the earth and its orbit is in the equatorial plane. If this satellite is not in the equatorial plane, it will appear to move up and down of the equatorial plane and thus it will not be stationary w.r.t. an observer on the Earth.

Example – 04:

With what velocity should a satellite be launched from a height of 600 km above the surface of the earth so as to move in a circular path?. Given: G = 6.67 x 10 -11 N m 2 /kg 2 radius of earth = 6.4 x 10 6 m, Mass of the earth = 5.98 x 10 24 Kg.

Given: G = 6.67 x 10 -11 N m 2 /kg 2 radius of earth = R = 6.4 x 10 6 m = 6400 km, Mass of the earth = M = 5.98 x 10 24 Kg, height of satellite above the surface of earth = 600 km, raius of orbit = 6400 + 600 = 7000 km = .7 x 10 6 m

To find: critical velocity = vc = ?

The critical velocity of a satellite orbiting around the earth is given by Ans: The velocity of the satellite for launching is 7.545 km/s.

Example – 05:

A satellite is revolving around the earth in a circular orbit at a distance of 10 7 m from its centre. Find the speed of the satellite. Given G = 6.67 x 10 -11 N m 2 /kg 2 , Mass of the earth = M =6 x 10 24 Kg,

Given: radius of orbit = 10 7 m, G = 6.67 x 10 -11 N m 2 /kg 2 , Mass of the earth = M = 6 x 10 24 Kg,

To Find: speed of satellite = vc =?

The critical velocity of a satellite orbiting around the earth is given by Ans: The speed of the satellite is 6.326 km/s.

Example – 06:

A body is raised to a height of 1600 km above the surface of the earth and projected horizontally with a velocity of 6 km/s. Will it revolve around the earth as a satellite? Given G = 6.67 x 10 -11 N m 2 /kg 2 radius of earth = 6.4 x 10 6 m, Mass of the earth = 6 x 10 24 Kg.

Given: radius of earth = 6.4 x 10 6 m = 6400 km, Mass of the earth = 6 x 10 24 Kg, height of satellite above the surface = h = 1600 km, horizontal velocity given to the satellite = v = 6 km/s, r = R + h = 6400 + 1600 = 8000 km, r = 8 x 10 6 m .

To find: the condition of orbiting of satellite = ?

The critical velocity of a satellite orbiting around the earth is given by The horizontal velocity given to satellite is 6 km/s which is less than the critical velocity of 7.073 km/s. Hence the satellite will not revolve around the earth in circular orbit but will fall back on the earth in a parabolic path.

Example – 07:

A body is raised to a height equal to the radius of the earth above the surface of the earth and projected horizontally with a velocity 7 km/s. Will it revolve around the earth as a satellite? If yes what is the nature of the orbit? Given G = 6.67 x 10 -11 N m 2 /kg 2 radius of earth = 6.4 x 10 6 m, Mass of the earth = 5.98 x 10 24 Kg.

Given: radius of earth = 6.4 x 10 6 m = 6400 km, Mass of the earth = 6 x 10 24 Kg, height of satellite above the surface = h = R, horizontal velocity given to the satellite = v = 7 km/s, r = R + R = 2R = 2 x 6400 = 12800 km, r = 12.8 x 10 6 m .

To find: the nature of the orbit of satellite =?

The critical velocity of a satellite orbiting around the earth is given by The horizontal velocity given to satellite is 7 km/s which is greater than the critical velocity 5.582 km/s and less than the escape velocity (11.2 km/s). Hence the satellite will revolve around the earth in an elliptical orbit.

Example – 08:

An artificial satellite is revolving around the earth in circular orbit at a height of 1000 km at a speed of 7364 m/s. Find its period of revolution if R = 6400 km

Given: Radius of earth =R = 6400 km = 6.4 x 10 6 m, height of satellite above the surface of the earth = h = = 1000 km = 1.0 x 10 6 m, radius of orbit of satellite = r = R + h = 6.4 x 10 6 + 1.0 x 106 m = 7.4 x 10 6 m, critical velocity = vc =7364 m/s.. Ans: The period of revolution of the satellite is 1.75 hr.

Example – 09:

Moon takes 27 days to complete one revolution around the earth. Calculate its linear velocity. If the distance between the earth and the moon is 3.8 x 10 5 km.

Given: Radius of orbit of moon = r = 3.8 x 10 5 km =3.8 x 10 8 m, Period of moon = T = 27 days = 27 x 24 x 60 x 60 s.

To find: critical velocity = vc =? Ans: The linear velocity of the moon is 1.023 km/s

Example – 10:

Find the radius of the moon’s orbit around the earth assuming the orbit to be circular. Period of revolution of the moon around the earth = 27.3 days, g at the earth’s surface = 9.8 m/s 2 . Radius of earth = 6400 km.

Given: Radius of earth = R = 6400 km = 6.4 x 10 6 m, Time period = T = 27.3 days = 27.3 x 24 x 60 x 60, g = 9.8 m/s 6 .

To Find: radius of the orbit = r =?  Ans: The radius of the moon’s orbit is 3.841 x 10 5 km

Example – 11:

Venus is orbiting around the sun in 225 days. Calculate the orbital radius and speed of the planet. Mass of sun is 2 x 10 30 kg, G = 6.67 x 10 -11 S.I. units.

Given: Mass of sun = M = 2 x 10 30 kg, Period of venus = T = 225 days = 225 x 24 x 60 x 60 s, G = 6.67 x 10 -11 S.I. units.

To Find: radius of orbit = r =? Orbital velocity = vc =?   Ans: Orbital radius of Venus = 1.085 x 10 11 m and its orbital speed = 3.506 x 10 4 m/s

Example – 12

An observer situated at the equator finds that a satellite is always overhead. What must be its distance from the centre of the earth? Given g at earth’s surface = 9.81 m/s 2 radius of earth = 6.4 x 10 6 m. What is the KE of such a satellite w.r.t. an observer on earth? What must be its height above the surface of the earth?

Given: Period of Earth = T = 24 hr = 24 x 60 x 60 s, Radius of the cearth = R = 6.4 x 10 6 m, g = 9.8 m/s 2 .

To Find: radius of the orbit = r =?  Hence, h = r – R = 42350 – 6400 = 35950 km

Ans: Radius of the orbit of the satellite is 42350 km. The height of the satellite above the surface of the earth is 35950 km.

As the satellite is stationary w.r.t. observer, the kinetic energy of satellite w.r.t. the observer is zero.

Example – 13:

Calculate the height of the communication satellite above the surface of the earth? Given G = 6.67 x 10 -11 N m 2 /kg 2 radius of earth = 6.4 x 10 6 m, Mass of the earth = 6 x 10 24 Kg.

Given: Period of satellite = T = 24 hr = 24 x 60 x 60 s, G = 6.67 x 10 -11 N m 2 /kg 2 radius of earth = R = 6.4 x 10 6 m, mass of earth = M = 6 x 10 24 Kg

To Find: height of satellite above the surface of the earth = h =?  Hence, h = r – R = 42.35 x 10 6 – 6.4 x 10 6 = 35.92 x 10 6 m

h = 35920 x 10 6 m = 35920 km

Ans: The height of satellite above the surface of the earth is 35920 km.

Example – 14:

A satellite makes ten revolutions per day around the earth. Find its distance from the earth assuming that the radius of the earth is 6400 km and g at the earth’s surface is 9.8 m/s 2 .

Given: radius of earth = R = 6400 km = 6.4 x 10 6 m, g = 9.8 m/s 2 , Number of revolutions = 10 per day

To Find: radius of orbit of satellite = r =?

T = 24/No. of revolutions per day = 24/10 = 2.4 hours  Ans: The radius of the orbit of the satellite is 9125 km

Example – 15:

A satellite is revolving around a planet in a circular orbit with a velocity of 8 km/s at a height where the acceleration due to gravity is 8 m/s 2 . How high is the satellite from the planet’s surface? Radius of planet = 6000 km.

Given: velocity of satellite vc = 8 km/s, R = 6400 km = 6.4 x 10 6 m, acceleration due to gravity at height = gh = 8 m/s 2 .

To Find: height of the satellite above the surface = h = ?,

The critical velocity of a satellite orbiting around the earth is given by Thus h = r – R = 8000 – 6400 = 1600 km

Ans: The height of the satellite above the surface of the earth is 1600 km.

Example – 16:

The critical velocity of a satellite is 5 km/s. Find the height of satellite measured from the surface of the earth given G = 6.67 x 10 -11 Nm 2 kg -2 R=6400 km and M = 5.98 x 10 24 kg

Given: critical velocity = vc = 5 km/s, 5 x 10 3 m, radius of earth = R = 6400 km = 6.4 x 10 6 m, G = 6.67 x 10 -11 Nm 2 kg -2 mass of earthM = 5.98 x 10 24 kg

To find: height of satellite above the surface of the earth = h = ?,

The critical velocity of a satellite orbiting around the earth is given by Thus h = r – R = 15950 – 6400 = 9550 km

Ans: The height of satellite above the surface of the earth is 9550 km.

Example – 17:

A satellite is revolving around a planet in a circular orbit with a velocity of 6.8 km/s. Find the height of the satellite from the planet’s surface and the period of its revolution. g = 9.8 m/s 2 , R = 6400 km.

Given: velocity of satellite = vc = 6.8 km/s = 6.8 x 10 3 m/s, R = 6400 km = 6.4 x 10 6 m, g = 9.8 m/s 2 .

To find: height of the satellite above the surface = h =?

The critical velocity of a satellite orbiting around the earth is given by Thus h = r – R = 8681 – 6400 = 2281 km Ans: The height of the satellite above the surface of the earth is 2281 km and the period of the satellite is 8017 s

## Calculation of Eccentricity of orbit from velocity and radius - Astronomy

Hohmann Transfers

A Hohmann Transfer is an orbital maneuver that transfers a satellite or spacecraft from one circular orbit to another. It was invented by a German scientist in 1925 and is the most fuel efficient way to get from one circular orbit to another circular orbit. Because the Hohmann Transfer is the most fuel efficient way to move a spacecraft, it is a fairly slow process and is used mostly for transferring spacecraft shorter distances.

A Hohmann Transfer is half of an elliptical orbit (2) that touches the circular orbit the spacecraft is currently on (1) and the circular orbit the spacecraft will end up on (3). It takes two accelerations to get the original orbit to the destination orbit. To move from a smaller circular orbit to a larger one the spacecraft will need to speed up to get onto the elliptical orbit at the perigee and speed up again at the apogee to get onto the new circular orbit. To move from a larger circluar orbit to a smaller one, the processes are reversed. In order to understand the Hohmann Transfer, students must first know a little about orbits. The Hohmann Transfer consists of two circular orbits and one elliptical orbit.

For circular orbits students should know: Discuss the radius of the orbit with students. Is the radius constant or does it change and why?

Students need to understand the difference between speed and velocity.

Speed is a scalar quantity which refers to "how fast an object is moving." A fast-moving object has a highspeed while a slow-moving object has a low speed. An object with no movement at all has a zero speed.
Velocity is a vector quantity which refers to "the rate at which an object changes its position." Imagine a person moving rapidly - one step forward and one step back - always returning to the original starting position. While this might result in a frenzy of activity, it would result in a zero velocity.

* Speed is a scalar and does not keep track of direction velocity is a vector and is direction-aware.

There are two types of forces involved in orbits. F1 is the force that pushes down on the Earth (in this case) from the other object and F2 is the force between two objects in general. Discuss with students the speed of an object in a circular orbit (constant because the radius is constant) and the force of an object (also constant). F2 always equals a constant and is always equal to F1.

Now that students know this information, they can calculate the velocity of an object in a circular orbit.

Let F1 = F2 and solve for V. We get, Also, discuss the inverse relationship between the radius of the circular orbit and the velocity of the object. Ask them what happens to the velocity as the radius increases and when it decreases.

For elliptical orbits students should know: Again, disucss with the students the speed of an object in an elliptical orbit. The speed is not constant because the radius changes. When the radius is the greatest, at the Apogee, velocity will be at the minmum. When the radius is the smallest, at the Perigee, velocity is at the maximum.

All that is left is to find the velocity of an object in an elliptical orbit. This is where it gets a little tricky.

*** Teacher's, from this point on use your discretion on how much information you want to give your studnets. How much you lead them will depend on the level of your students.

In order to find the velocity, students need to have a basic understanding of kinetic energy and potential energy. Potential energy is stored energy, while kinetic energy is energy of motion. It is the energy it possesses because of its motion. If we subtract an objects potential energy from its kinetic energy, we get the total amount of energy an object possesses. We know the formula for the potential energy, kinetic energy and the total energy of an object, so we can substitute those into the formula and solve for velocity. To simplify things, substitute , where a is half the distance of the major axis or the distance from one vertex to the center.

Now we know how to find the velocity of an object in a circular orbit and an elliptical orbit. Ask students what else they think they need to know to successfully make a Hohmann Transfer. Talk a little more about what it is and how it is done to lead them to the conclusion that they also need to know the velocity at the apogee and perigee in order to transfer the spacecraft from one orbit to the next.

In order to find the velocity at the apogee and perigee, it is important that students understand the eccentricity of ellipses and how to label a and c on an ellipse. Have students refer to the diagram below. We already know that the velocity of an object in a elliptical orbit is In order to find the velocity at A and P, we need to put the formula in terms of A and P. This is where eccentricity and our diagram come into play. Talk about whether velocity is faster at the apogee or perigee. Students should come to the conclusion that Vp > Va because the radius is the shortest at the perigee, which means velocity is at its fastest.

Now that students can find the velocity of an object in a circular and elliptical orbit, and the velocity of an object at the apogee and perigee of an elliptical orbit, they can begin to explain how to move a spacecraft from one circular orbit to another.

The following worksheet , adapted from Andrew Izsak's EMAT 6550 course on conic sections at UGA, will lead them in performing a Hohmann Transfer.

## Terms.

Feel free to browse through this list of terms at your leisure. Or, if you're looking for something specific, simply start typing in the search box.

#### Angular Momentum

The product of the amount by which a spinning object resists a change of its spin rate, or its direction of spin. This can be described mathematically by: $ar =I ar< Omega >$.

#### Angular Velocity

An object’s rate of spin. Also referred to as the angular velocity vector, or, $ar$.

#### Apogee

The point in the orbit of a celestial body (or satellite) at which it is furthest from Earth (i.e., the point of the largest radial distance in the geocentric-equatorial coordinate system).

#### Argument of Latitude

An angle, $u$, measured in the direction of a satellite’s motion, along the orbital path from the ascending node to the satellite’s position.

#### Geosynchronous Orbit

An inclined orbit with a period of approximately $24$ hours.

#### Gravity

Basically, the tendency of two or more objects to attract one another.

#### Gravity-Assist

Using a celestial body’s gravitational field and orbital velocity to “sling shot” a spacecraft ultimately resulting in changing the spacecraft’s velocity with respect to the Sun.

#### Great Circle

Any circle that slices through the center of a sphere.

#### Greenwich Mean Time (GMT)

The mean solar time at Greenwich, UK.

#### Heliocentric Coordinate System

A coordinate system with the center of the Sun as its origin.

#### Heliocentric-Ecliptic Coordinate System

Used for interplanetary transfer, this coordinate system’s origin in the center of the Sun, its fundamental plane is the ecliptic plane (see ecliptic plane), and its principal direction is in the direction of the vernal equinox.

#### Inclination

A Measurement that describes the tilt of an orbital plane with respect to its fundamental plane.

#### Inclination Auxiliary Angle

An angle expressed as $alpha$, and defined at the ascending node between the equator and the ground trace of an orbit.

#### Indirect Orbit

Also referred to as a retrograde orbit, this is an orbit in which a satellite moves opposite of Earth’s rotation, and has an inclination between $90^$and $180^$.

#### Inertia

A property of matter by which it remains in a state of rest or uniform motion until an external force is applied.

#### Kinetic Energy

The energy possessed by an object due to its motion.

#### Launch Window

The period of time when a spacecraft can be launched directly into a specific orbit from its launch site.

#### Launch-Direction Auxiliary Angle

An angle, $gamma$, measured at the intersection of an orbit’s ground trace and longitude line.

#### Launch-Window Location Angle

An angle expressed by the symbol $delta$, measured along the equator, between the node closest to the launch opportunity being considered and the longitude where the orbit crosses the launch site latitude.

#### Launch-Window Sidereal Time (LWST)

Measured from the direction of the vernal equinox to the point where the launch site passes through the orbital plane.

The angular distance, $alpha_$, that the target spacecraft travels during the intercepting spacecraft’s time of flight.

#### Line of Nodes

The two points that mark the intersection of the orbital plane and fundamental plane in a coordinate system.

#### Linear Momentum

The product of an object’s mass and its velocity. Mathematically speaking, this can be defined by the equation: $ar =m ar$.

#### Local Sidereal Time (LST)

The time since the vernal equinox passed over a particular (local) longitude line.

#### Longitude of Perigee

Represented by $Pi$, the angle measured from the principal direction to perigee, in the direction of a satellite’s motion.

A measurement of how much matter an object contains.

#### Mean Anomaly

Expressed as $M$, this is an angle that must be expressed mathematically (the little meanie has no physical meaning--it’s an angle in an imaginary orbit that corresponds to a celestial body’s eccentric anomaly). Mathematically, it is a product of a spacecraft’s mean motion and the amount of time that has passed since the spacecraft’s last perigee passage, or: $M = nT$.

#### Mean Motion

Expressed as $n$, the angular speed of a spacecraft.

#### Mean Solar Day

The average time between the Sun’s successive passages over a given longitude.

#### Molniya Orbit

A semi-synchronous, eccentric orbit used for specific communications satellites (particularly those providing coverage in northern latitudes, i.e., Russia, Canada, etc.).

#### Moment Arm

The moment arm, or $ar$, is the measurement of the perpendicular length between an joint axis and the line of force acting upon that joint.

#### Moment of Inertia

An object’s resistance to spin, represented mathematically as $I$.

#### Momentum

The amount of resistance possessed by an object in motion.

#### Nodal Displacement

A measurement (represented as $Delta N$) of how much an orbit’s ground track moves to the west from one orbit to the next.

#### Orbit

A fixed path on which a spacecraft (or anything, really) travels around a planet or other celestial body.

#### Orbit Cranking

A technique used to change the direction of a spacecraft’s velocity.

#### Orbit Pumping

A technique used to change the magnitude of a spacecraft’s velocity.

#### Origin

A physically identifiable starting point for a coordinate system.

#### Parking Orbit

A temporary orbit used by a spacecraft until it transfers to its final mission orbit.

#### Patched-Conic Approximation

A simplifying assumption that breaks the interplanetary trajectory into three separate regions, and only takes the gravitational attraction between the spacecraft and one celestial body in each region into consideration. (A major assumption of this approximation is that all orbital planes lie in the ecliptic plane).

#### Perigee

The point in the orbit of a celestial body (or satellite) at which it is closest to Earth (i.e., the point of the smallest radial distance in the geocentric-equatorial coordinate system).

#### Polar Orbit

An orbit that follows a path between the North and South Poles, having an inclination of $90^$.

#### Potential Energy

The energy possessed by an object due to its relative position to other objects.

#### Principal Direction

Defined by pointing a unit vector toward a visible, distant object, e.g., the north star.

The largest radial distance of an ellipse. In a geocentric-equatorial coordinate system, this is called apogee.

The smallest radial distance of an ellipse. In a geocentric-equatorial coordinate system, this is referred to as perigee.

#### Range

A satellite’s (or spacecraft’s) distance from its tracking site.

#### Re-Entry Coordinate System

A coordinate system with the spacecraft’s center of mass (at the start of re-entry) as its origin, the spacecraft’s orbital plane as its fundamental plane, and “down,” or the Earth’s center as its principal direction.

#### Re-Entry Corridor

The three dimensional path a spacecraft must follow when entering Earth’s atmosphere, to avoid skipping out or burning up.

#### Rendezvous

The arrival of two or more spacecraft at the same point in an orbit at the same time.

#### Right Ascension of the Ascending Node

Represented by $Omega$, an angle describing orbital orientation with respect to the principal direction.

#### Semi-Synchronous Orbit

An orbit with an orbital period of $12$ hours.

#### Semimajor Axis

One half the distance across the long axis of an ellipse.

#### Sidereal Day

The time between passages of the vernal equinox over a given longitude.

#### Sphere of Influence (SOI)

The volume of space within which a celestial body’s gravitational force dominates.

#### Sun-Synchronous Orbit

A retrograde (indirect), low Earth orbit, with an inclination ranging between $95^$ and $105^$. This orbit is often used for remote sensing satellites.

#### Swath-Width

The linear width, or diameter, of specific total area that is visible (typically, as seen by a sensor) on Earth's surface at one time.

#### Topocentric-Horizon Frame

An Earth-fixed reference frame with its origin at the launch site and the horizontal as its fundamental plane.

#### Total Mechanical Energy

A measurement resulting from the sum of an object’s position, or potential energy, and its motion, or kinetic energy. Mathematically, this can be defined as $E = PE + KE$.

#### Trajectory

The path an object follows through space.

#### Transfer Orbit

An intermediate orbit that transfers a spacecraft from its parking orbit to its final mission orbit.

#### Two-Line Element Set (TLE)

A data format, consisting of two 69-character lines, used to encode a list of orbital elements of an Earth-orbiting satellite.