# Homework Help: Basic Orbital Problem

1. Oct 20, 2011

### Kyle91

1. The problem statement, all variables and given/known data

If the Apollo vehicle is initially in a circular low Earth orbit with altitude h0 = 320 km
and it is suddenly accelerated to a new (tangential) velocity of 10.8 km/s, determine the
radial position, r, at apogee, the eccentricity of the new orbit and the period of this orbit.

---- I was confused by the wording but just to be clear it starts in a circular orbit and ends in an elliptical orbit of velocity 10.8km/s.

2. Relevant equations

Eccentricity: e = (rap - rper)/(rap + rper)

3. The attempt at a solution

I think there must be a relation between starting in a circular orbit and ending in an elliptical one. Is it possible that the nearest the Apollo vehicle will come to the Earth under this orbit is 320km? And only the aphogee requires calculation?

I'm mainly asking this to see if anyone knows the relationship between the Mass of Earth (negligible Apollo mass) and velocity of orbit. I've had a quick google but unfortunately can't find it!

Thanks!

Last edited: Oct 20, 2011
2. Oct 20, 2011

### Kyle91

3. Oct 20, 2011

### ehild

You need to use the following laws: Conservation of angular momentum, conservation of energy, and that the centripetal force for a circular orbit or radius r is equal to the force of gravity at distance r form the centre of the Earth. The mass of Earth is 1 M⊕ = 5.9722 × 1024 kg.

ehild

4. Oct 20, 2011

### Kyle91

But the orbit isn't circular! Do I assume the perigee is 320km?

5. Oct 20, 2011

### vela

Staff Emeritus
What's special about the velocity of an object when it's at the perigee or apogee of its orbit?

6. Oct 20, 2011

### Kyle91

Perigee is the point of the orbit where the object is closest to the body it's orbiting and the apogee is when it's furthest :)

7. Oct 20, 2011

### vela

Staff Emeritus
But what about its velocity? What characteristic is unique to those two points?

8. Oct 20, 2011

### Kyle91

Equal and opposite. If you think about it in an xy-plane where both the perigee and apogee are at x = 0 then dy/dt = 0 at the points also (I think).

What I'm trying to say if you picture the orbit like this: http://www.gap-system.org/~history/Curvepics/Ellipse/Ellipse1.gif [Broken], then at the left and right points the velocity in the x direction is zero.

...hmm, this would mean all the velocity/kinetic energy is in the y direction. Am I on the right track?

Last edited by a moderator: May 5, 2017
9. Oct 20, 2011

### vela

Staff Emeritus
In other words, the velocity is perpendicular to the radial vector at those two points. The radial velocity is 0 because at apogee and perigee, r is at a maximum or minimum, so the velocity is completely tangential.

You should be able to answer the question you asked in post 4.

10. Oct 20, 2011

### Kyle91

Could you please explain the radial vector to me? The velocity is always tangential to the radial vector throughout the body's entire orbit, right?

I thought the radial vector was the position vector (or gravitational force vector) from Earth to the shuttle. In that case its magnitude should never be zero.

11. Oct 20, 2011

### vela

Staff Emeritus
The radial vector r indeed goes from the center of the Earth to the vehicle. Its tip traces out the orbit. The velocity vector is always tangential to the orbit, but that doesn't mean it's perpendicular to r.

Note I said the radial velocity is 0 at the apogee and perigee. I'm not saying r equals 0 at those points.

12. Oct 20, 2011

### Kyle91

I haven't heard the term radial velocity before, but after a quick wikipedia I think it means the velocity towards the orbiting body. Which makes sense that it'd be zero as it's essentially the turning point. The point at which the sign of the radial velocity changes.

That said, I don't know how this helps my question posed in #4 :S

(I'm really trying not to make you just give me the answer - nor do I want you to unless there's a formula - sorry if it seems that way)

13. Oct 20, 2011

### vela

Staff Emeritus
Originally, the orbit was circular, so the velocity is perpendicular to the orbit. Therefore...

14. Oct 20, 2011

### Kyle91

Therefore the perigee is 320km. Thanks.

Is there an easy formula now for finding the apogee? (That you know off of the top of your head, I'm having a google!)

15. Oct 20, 2011

### vela

Staff Emeritus
Not that I know of. Use ehild's suggestion. It's a pretty straightforward problem.

16. Oct 20, 2011

### logics

try this: http://wikipedia.org/wiki/Orbital_speed" [Broken]

Last edited by a moderator: May 5, 2017
17. Oct 20, 2011

### Kyle91

Thanks logics!

I was actually reading something else just before, and I reached another conundrum. What is the velocity I'm quoted? Is that the mean orbital velocity or would it be the velocity at perigee? I have an annoying feeling it's the mean velocity but I'm not sure :/

18. Oct 20, 2011

### Kyle91

...or alternatively am I meant to assume the change in velocity is negligible?

19. Oct 20, 2011

### ehild

The orbit is elliptical, but the velocity is perpendicular to the radius of the orbit at both perigee and apogee. From the conservation of angular momentum, mv1R1=mv2R2. You know the speed and radius at the perigee, the product of the speed and radius is the same at the apogee. But take care, the radius is the distance from the centre of Earth, and the altitude is the hight above the Earth surface so r1=Rearth+320 km.
Conservation of energy provides an other relation between velocity and radius at apogee. What is the potential energy at distance r from the Earth?

ehild

20. Oct 20, 2011

### Kyle91

But how do I find v1? Or is that the 10.8km/s? I thought that'd be the mean orbital velocity :S

To find the potential energy at distance r, I'd first find the potential energy at perigee --> U = mgh = m*9.8*r1

Then the kinetic energy --> 0.5*m*v12

Sum them = m(9.8*r1 + 0.5*v12)

And then take away the kinetic energy at distance r:

m(9.8*r1 + 0.5*v12) - 0.5*m*v

21. Oct 20, 2011

### ehild

v1=10.8 km/s.

Uhhh. It is the potential energy with respect to the surface of Earth and very close to the surface. You need to use the gravitational potential energy, corresponding to the universal law of gravity.
What is the force between two masses at distance R?

ehild

22. Oct 20, 2011

### Kyle91

*Sigh* Thank you for being patient, I've been working on this seriously all day. And it's just been one of those days.

G = U*m1*m2/r^2

Or in this case G = U*m/r^2

23. Oct 20, 2011

### ehild

The force of gravity is Gm1m2/r^2, and G is the gravitational constant.
The gravitational potential around the Earth is -GM/r.
From conservation of energy, 1/2 v12-GM/r1=1/2 v22-GM/r2, or
1/2 (v12-v22)=GM(1/r1-1/r2).

ehild

Last edited: Oct 20, 2011