Calculate the orbital velocity in a geostationary orbit

In summary: I can use their approximation of R=40000km... the first step is to show this approximation is valid, so you cannot use it.He did show that the approximation is valid. He arrived at a value of 42,000 km, which is approximately 40,000 km. I think you missed the point of fayled's question.Here's what he's asking: He has three options,Calculate the delta v needed to circularize a 8000 km by 42164 km at 42164 km using the correct values for those two orbits,Calculate the delta v needed to circularize a 8000 km by 40000 km at 40000 km using the correct values
  • #1
fayled
177
0

Homework Statement


Calculate the orbital velocity in a geostationary orbit (the circular orbit around
the Earth which has a period of 24 hours) and show that its radius is approximately
40,000 km.

A satellite is to be inserted into a geostationary orbit from an elliptical orbit with perigee at a geocentric radius of 8,000 km and apogee at 40,000 km. When it is at apogee, a brief firing of its rocket motor places it into the circular orbit. Calculate the change in velocity the motor needs to provide.

Homework Equations


Conservation of angular momentum and of energy.
L=mr2(dθ/dt), L=mrxv.
E=m(dr/dt)2/2+J2/2mr2-GmM/r
where M is the mass of the Earth, m that of the satellite.

The Attempt at a Solution


In the geostationary orbit, I have v=3076ms-1 and R=42000km. Now the first issue is whether or not I can use their approximation of R=40000km and so then assume that the elliptical orbit has the correct radial distance, just not the correct speed at apogee.

Assuming I can do that, equating the energy at apogee and perigee using dr/dt=0 and using r=rmin at perigee and r=rmax at apogee, I can find L2. Then using L2=m2rmax2v2 at apogee gives v=sqrt{2GMrmin/[rmax(rmin+rmax)]}=1826ms-1.

This concerns me because the energy of an elliptical orbit is greater than that of a circular one. At the position in consideration, they have the same PE, they both have no radial KE contribution, so comparing their tangential KE compares their overall energies. My result that the circular orbit has a greater tangential velocity seems to suggest that the circular orbit has a greater energy. So this must be wrong unless I'm misunderstanding something.

For the sake of completion, the speed boost needed is then an increase of 1250ms-1. Thanks for any help with these issues in advance.
 
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  • #2
the first issue is whether or not I can use their approximation of R=40000km
... the first step is to show this approximation is valid, so you cannot use it.
But, once you've shown the approximation is valid, you can probably get away with using the round figure to get approximate relationships for the elliptical orbit.

When you talk about energy - be careful to say what form of energy.
What leads you to say that the energy of an elliptical orb it is greater than that for a circular one?
Is this always the case?
 
  • #3
Simon Bridge said:
... the first step is to show this approximation is valid, so you cannot use it.
He did show that the approximation is valid. He arrived at a value of 42,000 km, which is approximately 40,000 km. I think you missed the point of fayled's question.

Here's what he's asking: He has three options,
  • Calculate the delta v needed to circularize a 8000 km by 42164 km at 42164 km using the correct values for those two orbits,
  • Calculate the delta v needed to circularize a 8000 km by 40000 km at 40000 km using the correct values for those two orbits, or
  • Calculate the delta v needed to circularize a 8000 km by 40000 km at 40000 km, but this time using the correct velocity for the transfer orbit but the orbital velocity for a circular orbit at 42164 km.
Which to use? The first choice is the correct choice. This however does not appear to be what the question implies fayled should be doing. It appears to me the question is asking that the third choice be used. The error induced by using the second choice is rather small, about 4 m/s. The error induced by using the third choice is much larger, about 90 m/s.

However, fayled's 1250m/s result for this choice is also "wrong".

fayled, that 40,000 km figure entails rounding to **one** significant digit. You should do much the same for your delta v. You definitely should not be reporting things three significant digits.


fayled said:
This concerns me because the energy of an elliptical orbit is greater than that of a circular one.
No! The total mechanical energy of an satellite with mass m orbiting the Earth with semi-major axis a is ##-\frac{GMm}{2a}##. Eccentricity plays no part. Increasing the semi-major axis increases the energy. This maneuver is increasing the semi-major axis, from 24,000 km to 40,000 km.
 
  • #4
D H said:
He did show that the approximation is valid. He arrived at a value of 42,000 km, which is approximately 40,000 km. I think you missed the point of fayled's question.

Here's what he's asking: He has three options,
  • Calculate the delta v needed to circularize a 8000 km by 42164 km at 42164 km using the correct values for those two orbits,
  • Calculate the delta v needed to circularize a 8000 km by 40000 km at 40000 km using the correct values for those two orbits, or
  • Calculate the delta v needed to circularize a 8000 km by 40000 km at 40000 km, but this time using the correct velocity for the transfer orbit but the orbital velocity for a circular orbit at 42164 km.
Which to use? The first choice is the correct choice. This however does not appear to be what the question implies fayled should be doing. It appears to me the question is asking that the third choice be used. The error induced by using the second choice is rather small, about 4 m/s. The error induced by using the third choice is much larger, about 90 m/s.

However, fayled's 1250m/s result for this choice is also "wrong".

fayled, that 40,000 km figure entails rounding to **one** significant digit. You should do much the same for your delta v. You definitely should not be reporting things three significant digits.



No! The total mechanical energy of an satellite with mass m orbiting the Earth with semi-major axis a is ##-\frac{GMm}{2a}##. Eccentricity plays no part. Increasing the semi-major axis increases the energy. This maneuver is increasing the semi-major axis, from 24,000 km to 40,000 km.

If we consider the effective potential plot as a function of radia position, then the circular orbit has the minimum energy (corresponding to one radial position) and the elliptical orbit has some larger energy so that it has two 'radial limits', at least that's what I have been taught. Is this not right?
 
  • #5
fayled said:
If we consider the effective potential plot as a function of radia position, then the circular orbit has the minimum energy (corresponding to one radial position) and the elliptical orbit has some larger energy so that it has two 'radial limits', at least that's what I have been taught. Is this not right?

You're writing about this plot:

activities:content:photos:cfeffpotential.jpg


That's a useful but potentially misleading diagram. It does not mean what you think it means. In particular, it does not mean what you wrote in the opening post, "This concerns me because the energy of an elliptical orbit is greater than that of a circular one."

Imagine a particle orbiting circularly far removed from the central body and another particle orbiting circularly close into the central body. That remote particle has much greater mechanical energy than does the close-in particle. The mechanical energy of the remote particle is almost zero; give it just a tiny nudge forward and it will escape. The close-in particle has a large negative mechanical energy. Give it a tiny nudge forward and it will now have a slightly elliptical orbit. That tiny nudge was a just tiny nudge. It didn't suddenly gain so much energy that it exceeds that of the remote particle.

That red curve is the effective potential energy for a given angular momentum. A different angular momentum results in a different effective potential energy curve.

Somewhere in your text or your notes you should have a derivation for the total mechanical energy for an orbiting object. It should be along the lines of ##E=-\frac {GMm}{2a}##, where ##a## is the semi-major axis of the orbit.
 
  • #6
D H said:
You're writing about this plot:

activities:content:photos:cfeffpotential.jpg


That's a useful but potentially misleading diagram. It does not mean what you think it means. In particular, it does not mean what you wrote in the opening post, "This concerns me because the energy of an elliptical orbit is greater than that of a circular one."

Imagine a particle orbiting circularly far removed from the central body and another particle orbiting circularly close into the central body. That remote particle has much greater mechanical energy than does the close-in particle. The mechanical energy of the remote particle is almost zero; give it just a tiny nudge forward and it will escape. The close-in particle has a large negative mechanical energy. Give it a tiny nudge forward and it will now have a slightly elliptical orbit. That tiny nudge was a just tiny nudge. It didn't suddenly gain so much energy that it exceeds that of the remote particle.

That red curve is the effective potential energy for a given angular momentum. A different angular momentum results in a different effective potential energy curve.

Somewhere in your text or your notes you should have a derivation for the total mechanical energy for an orbiting object. It should be along the lines of ##E=-\frac {GMm}{2a}##, where ##a## is the semi-major axis of the orbit.

So, analysing the effective potential curve:
The elliptical orbit has a smaller angular momentum (as I calculated above) - incidentally is there a more obvious way of seeing that an elliptical orbit of semi major axis R has a smaller L than a circular orbit of radius R without finding the velocity at rmax?
So it's effective potential curve shifts downwards. Then it's maximum radial position has to coincide with the radius of the circular orbit (the trough needs to move left as well which the smaller L takes care of). This then implies their energies will be equal. This also agrees with your formula for the energy, so all is good in that sense.

However back now to my original scenario:
At rmax, both orbits have equal energy. dr/dt=0 for both, and the both have the same PE. This implies their tangential velocities must be equal so their energies are. However they are not as I calculated. So the situation still hasn't been fully resolved (maybe what I wrote above was wrong?).

Thanks.
 
  • #7
fayled said:
So, analysing the effective potential curve:
The elliptical orbit has a smaller angular momentum (as I calculated above) - incidentally is there a more obvious way of seeing that an elliptical orbit of semi major axis R has a smaller L than a circular orbit of radius R without finding the velocity at rmax?
Sure.

Note: I'm using specific angular momentum ##\ell## and specific orbital energy ##E##, angular momentum and energy divided by reduced mass. We haven't sent any large masses into orbit, so reduced mass is essentially the mass of the orbiting body. It makes the math so much easier not having to deal with reduced mass. I'm also using ##\mu_E## to denote the product ##GM_E##. This product is known to nine places. G (and hence the mass of the Earth) are only known to four places. It's much better to use the standard gravitational parameter ##\mu_E## than G*M.

Hopefully this isn't too confusing for you.

Method #1: The specific angular momentum for a satellite orbiting the Earth is ##\ell^2= \mu_E a (1-e^2)## where ##a## is the semi-major axis and ##e## is the eccentricity. For a given semi-major axis, this obviously takes on a maximum at zero eccentricity.

Method #2: Here I'll use the vis-viva equation, ##v^2 = \mu_e\left(\frac 2 r - \frac 1 a\right)## . Note that when ##r=a## this reduces to ##v^2 = \frac {\mu_E}{a}## . That's the circular orbit velocity for an orbit with that semi-major axis ##a##. No matter what the shape of the orbit, the magnitude of the velocity at ##r=a## is that of a circular orbit velocity at that radial distance. This means the angular momentum of an orbit for a given semi-major axis will be maximum when the velocity vector and radial vector at ##r=a## are orthogonal. That's only the case for a circular orbit.
However back now to my original scenario:
At rmax, both orbits have equal energy.
No, they don't. The first orbit has a semi-major axis of 24,000 km. The second, a semi-major axis of 40,000 km. Energy is a function of the gravitational parameter and semi-major axis only. The transfer orbit (the elliptical orbit that took the vehicle from 8,000 km to 40,000 km) has less (more negative) energy than does the outer circular orbit.

This is admittedly a bit confusing. (Some say more than a bit.) When the vehicle is orbiting circularly at 8,000 km it has to fire its thrusters to increase its velocity to put it on that transfer orbit. Once the vehicle reaches apogee, it has to fire its thrusters to increase its velocity once again to put it in that new circular orbit. The most confusing part: The vehicle is now moving slower than it was when it started orbiting circularly at 8,000 km.

I'm not making this up. Google the term "Hohmann transfer".

In many cases a Hohmann transfer is the most energy efficient means by which a vehicle can move from one circular orbit to another. But not in all cases! For something even more confusing, suppose you want to transfer from a close-in circular orbit with radius ##a## to a circular orbit with radius ##12a##. Now it's better to make two transfer half orbits. The first takes the vehicle from a distance of ##a## to a very, very large distance. The second brings the vehicle back to the target orbit. Finally, at the target distance it fires the jets one last time to circularize. This is a bi-elliptic transfer. But you don't have to worry about that for this course.
 
  • #8
fayled said:
A satellite is to be inserted into a geostationary orbit from an elliptical orbit with perigee at a geocentric radius of 8,000 km and apogee at 40,000 km. When it is at apogee, a brief firing of its rocket motor places it into the circular orbit. Calculate the change in velocity the motor needs to provide.

The Attempt at a Solution


In the geostationary orbit, I have v=3076ms-1 and R=42000km. Now the first issue is whether or not I can use their approximation of R=40000km and so then assume that the elliptical orbit has the correct radial distance, just not the correct speed at apogee.

Assuming I can do that, equating the energy at apogee and perigee using dr/dt=0 and using r=rmin at perigee and r=rmax at apogee, I can find L2. Then using L2=m2rmax2v2 at apogee gives v=sqrt{2GMrmin/[rmax(rmin+rmax)]}=1826ms-1.

This concerns me because the energy of an elliptical orbit is greater than that of a circular one. At the position in consideration, they have the same PE, they both have no radial KE contribution, so comparing their tangential KE compares their overall energies. My result that the circular orbit has a greater tangential velocity seems to suggest that the circular orbit has a greater energy. So this must be wrong unless I'm misunderstanding something.

The circular orbit of radius R has greater energy as the elliptic one with distance of apogee rmax =R. At apogee, the PE is the same on both orbits, but the velocities are not equal. The centripetal acceleration is also the same, GmM/R2=v2/ρ, where ρ is the radius of curvature. The elliptical orbit is more curved, its radius of curvature is smaller than the radius of the circular orbit. Therefore, the speed on the elliptical orbit must be smaller than that on the circular orbit.

ehild
 

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  • #9
D H said:
Sure.

Note: I'm using specific angular momentum ##\ell## and specific orbital energy ##E##, angular momentum and energy divided by reduced mass. We haven't sent any large masses into orbit, so reduced mass is essentially the mass of the orbiting body. It makes the math so much easier not having to deal with reduced mass. I'm also using ##\mu_E## to denote the product ##GM_E##. This product is known to nine places. G (and hence the mass of the Earth) are only known to four places. It's much better to use the standard gravitational parameter ##\mu_E## than G*M.

Hopefully this isn't too confusing for you.

Method #1: The specific angular momentum for a satellite orbiting the Earth is ##\ell^2= \mu_E a (1-e^2)## where ##a## is the semi-major axis and ##e## is the eccentricity. For a given semi-major axis, this obviously takes on a maximum at zero eccentricity.

Method #2: Here I'll use the vis-viva equation, ##v^2 = \mu_e\left(\frac 2 r - \frac 1 a\right)## . Note that when ##r=a## this reduces to ##v^2 = \frac {\mu_E}{a}## . That's the circular orbit velocity for an orbit with that semi-major axis ##a##. No matter what the shape of the orbit, the magnitude of the velocity at ##r=a## is that of a circular orbit velocity at that radial distance. This means the angular momentum of an orbit for a given semi-major axis will be maximum when the velocity vector and radial vector at ##r=a## are orthogonal. That's only the case for a circular orbit.



No, they don't. The first orbit has a semi-major axis of 24,000 km. The second, a semi-major axis of 40,000 km. Energy is a function of the gravitational parameter and semi-major axis only. The transfer orbit (the elliptical orbit that took the vehicle from 8,000 km to 40,000 km) has less (more negative) energy than does the outer circular orbit.

This is admittedly a bit confusing. (Some say more than a bit.) When the vehicle is orbiting circularly at 8,000 km it has to fire its thrusters to increase its velocity to put it on that transfer orbit. Once the vehicle reaches apogee, it has to fire its thrusters to increase its velocity once again to put it in that new circular orbit. The most confusing part: The vehicle is now moving slower than it was when it started orbiting circularly at 8,000 km.

I'm not making this up. Google the term "Hohmann transfer".

In many cases a Hohmann transfer is the most energy efficient means by which a vehicle can move from one circular orbit to another. But not in all cases! For something even more confusing, suppose you want to transfer from a close-in circular orbit with radius ##a## to a circular orbit with radius ##12a##. Now it's better to make two transfer half orbits. The first takes the vehicle from a distance of ##a## to a very, very large distance. The second brings the vehicle back to the target orbit. Finally, at the target distance it fires the jets one last time to circularize. This is a bi-elliptic transfer. But you don't have to worry about that for this course.

Ah, I was very silly, obviously the semi major axis of the ellipse is the distance from the centre of the ellipse not the focal point! Ok, I've come across that first method before. This probably tells me I should try and remember the geometrical relations, i.e between energy and eccentricity/angular momentum and eccentricty, which I've been a little ignorant of in the past.

And how do you know what's on my course :wink:.

There is a final part to the question:
Due to a programming error the motor fires when the satellite is at perigee instead
of apogee (in the sense as to increase the orbital velocity in the direction of motion),
with the same change in velocity as before. Does the resulting orbit have the same,
greater, or less energy than the geostationary orbit?

Answer: energy is conserved, so the satellite has the same energy at perigee and apogee. If at apogee the change in velocity gave the orbit enough energy to become circular, then the boost at perigee must increase the energy of the satellite by the same amount, so again the energy is the same as in the circular orbit. This is probably horribly wrong.

As an extra, the 'boost' at apogee gave for a circular orbit because it resulted in the angular momentum and energy being the same as that of the circular orbit. At perigee, this boost gives the same energy as the circular orbit, but not the same angular momentum, thus the orbit remains elliptical.

Thanks again for your help :)
 
  • #10
fayled said:
Now the first issue is whether or not I can use their approximation of R=40000km and so then assume that the elliptical orbit has the correct radial distance, just not the correct speed at apogee.
The problem only gives numbers to one significant figure (except for there being 24 hours in a day). Your answer, to one significant figure, is 40000 km.
 
  • #11
fayled said:
There is a final part to the question:
Due to a programming error the motor fires when the satellite is at perigee instead
of apogee (in the sense as to increase the orbital velocity in the direction of motion),
with the same change in velocity as before. Does the resulting orbit have the same,
greater, or less energy than the geostationary orbit?

Answer: energy is conserved, so the satellite has the same energy at perigee and apogee. If at apogee the change in velocity gave the orbit enough energy to become circular, then the boost at perigee must increase the energy of the satellite by the same amount, so again the energy is the same as in the circular orbit. This is probably horribly wrong.
Rather than telling you the complete answer, I will just say that one part of this answer is correct.

It's the "This is probably horribly wrong" part.

Hint: Expand (v+Δv)2.
 
  • #12
D H said:
Rather than telling you the complete answer, I will just say that one part of this answer is correct.

It's the "This is probably horribly wrong" part.

Hint: Expand (v+Δv)2.

You mean my 'as an extra' part is wrong right?

Well expanding gives v2+2vΔv+Δv2. Not sure where this is going though :(

(also sorry for the delay in reply)
 

Related to Calculate the orbital velocity in a geostationary orbit

What is a geostationary orbit?

A geostationary orbit is a type of orbit around the Earth in which a satellite appears to remain stationary over a fixed location on the Earth's surface. This is achieved by placing the satellite in an equatorial orbit at a specific altitude where its orbital period matches the Earth's rotational period.

How is the orbital velocity in a geostationary orbit calculated?

The orbital velocity in a geostationary orbit can be calculated using the formula v = √(GM/R), where v is the orbital velocity, G is the gravitational constant, M is the mass of the Earth, and R is the distance between the satellite and the center of the Earth.

What is the standard altitude for a geostationary orbit?

The standard altitude for a geostationary orbit is approximately 35,786 kilometers (22,236 miles) above the Earth's surface. This altitude is also known as the geostationary orbit height or the Clarke orbit.

Why is the orbital velocity in a geostationary orbit important?

The orbital velocity in a geostationary orbit is important because it allows the satellite to maintain a fixed position over a specific location on the Earth's surface. This is crucial for applications such as telecommunications, weather monitoring, and navigation.

Is the orbital velocity in a geostationary orbit the same as the Earth's rotational velocity?

No, the orbital velocity in a geostationary orbit is not the same as the Earth's rotational velocity. The Earth's rotational velocity varies depending on the latitude, while the orbital velocity in a geostationary orbit is constant at 3.07 kilometers per second (1.90 miles per second).

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