# Velocity of a satellite in an elliptical orbit

1. Oct 12, 2009

### ehilge

1. The problem statement, all variables and given/known data
see attachment #12.106

2. Relevant equations
V=R$$\sqrt{}(g/r)$$ (for a circular orbit)
where R is the radius of the earth and r is the radius of the orbit from the center of the earth

conservation of momentum for elliptical orbits:
Vara=Vbrb

3. The attempt at a solution
The first thing I did was find the velocity of the satellite while it is still in a circular orbit and came up with 1.46x108 m/s. Now this is fine and dandy but I don't see where there is enough information to get the velocities in the elliptical orbit since the only equation I have for an elliptical orbit is listed above and I don't have Va or Vb. I tried to pretend that the object also went into a circular orbit at B by another rocket boost. Hoping that I might be able to solve for something but to no avail. So I guess my question really is, what is another equation that also has Va and Vb so I can solve simultaneously, or is there a way to eliminate one of the variables that I don't see?
Thanks!

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2. Oct 12, 2009

### Andrew Mason

Angular momentum is conserved. So, $m\omega r^2 = constant$ or as you have noted, Vara=Vbrb.

You have Va and ra. What you need to find is rb - the effective radius at the farthest point. Would that not just be the distance from the centre of the earth? Be careful about translating radius from altitude.

AM

Last edited: Oct 12, 2009
3. Oct 12, 2009

### Staff: Mentor

This in general is not true for elliptical orbits. Conservation of momentum says $\mathbf r \times \mathbf v$ is constant. This is the vector cross product; the radial component of velocity is not involved in this expression. For elliptical orbits, the radial component of velocity is zero at two points: apogee and perigee. Thus $r_av_a = r_p v_p$ is valid.

4. Oct 12, 2009

### Andrew Mason

Yes. The angular momentum is the tangential component of velocity divided by r. The confusion is avoided if one uses $m\omega r^2$ for angular momentum.

AM

5. Oct 12, 2009

### ehilge

first off, I realize that the radius at the furthest point is the altitude above the earth + radius of the earth, and I have factored this into my calculations.

Secondly, I do not have Va of the object in the elliptical orbit, only in the original circular orbit. And I need to find the increase in speed the object makes at A while in a circular orbit to project it into an elliptical orbit. This is the part that I don't understand because I have two unknowns with the one equation.

ok, fair enough and good to know. However, the points I am considering in the problem are at apogee and perigee, so the original equation I had is valid, right? And I still have the problem of having one equation with two unknowns. Any ideas on how to resolve that?

6. Oct 12, 2009

### Staff: Mentor

Well there's always the conservation laws. You've already used conservation of angular momentum. What else is conserved in an orbit?

BTW, beware that the question used altitudes, not distances from the center of the Earth. You might find the latter to be a better choice. For example, in $r_av_a = r_p v_p$.

7. Oct 13, 2009

### Andrew Mason

What is the change in potential energy of the satellite in moving between A and B? Does that help you determine the relationship between speeds at A and B?

AM

8. Oct 13, 2009

### ehilge

I managed to find another equation that I can use

1/rapogee+1/rperigee=2GM/h2 = 2(gR2)/(raVa)2 = 2(gR2)/(rbVb)2

I used this to find the velocity at A, and then calculated everything else from there using the previous equations.

I suppose I could also have used conservation of energy though

KEA + GPEA = KEB + GPEB

.5mvA2 + mgha = .5mvB2 + mghB

mass cancels out and I have another equation with va and vb

9. Oct 13, 2009

### Andrew Mason

You have to explain where this formula comes from. Consider:

$$\Delta KE + \Delta PE = 0$$

$$1/2(mv_A^2 - mv_B^2) + (-GMm/r_A + GMm/r_B) = 0$$

$$v_A^2 - v_B^2 = 2(GM/r_A - GM/r_B)$$

From the conservation of angular momentum:

$$v_Ar_A = v_Br_B$$

so you can solve for vA

AM