# Homework Help: Orbit Question

1. Jan 27, 2008

### GreenLRan

[SOLVED] Orbit Question

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

A satellite is in a circular earth orbit of altitude 400km. Determine the new perigee and apogee altitudes if the satellite on-board engine gives the satellite a radial (outward) component of velocity of 240m/s.

Answers: Z_apogee = 621km, Z_perigee = 196km

2. Relevant equations
mu = 398600km^3/sec^2
h (specific angular momentum) = r*v_perpendicular
a = semimajor axis of ellipse
v^2/2 + mu/r = -mu/(2*a) ( specific energy equation )
e= eccentricity of the orbit
theta = the true anomaly angle ( angle between eccentricity vector and the position vector)
r_perigee = a(1-e)
radius of the earth = 6378km

r = h^2/mu * (1+e*cos(theta))

V_cir = sqrt(mu/r)

3. The attempt at a solution

I started off by getting the magnitude of the velocity (v_perp^2 + v_radial^2)^.5
then calculated the escape velocity to see if it would be a closed orbit.

I found that it was still an ellipse (i also calculated the specific energy and found that it was negative... thus being an ellipse) from the specific energy i got 'a' ( -mu/(2*a) )
and using rp = a(1-e) = h^2/mu*(1+e)^-1 (at theta = 0 for rp)

however... when i did this I used the same h that i got from the satellite being a circular orbit.. which i am unsure if that was incorrect to do..

finally after solving for e, i plugged it back into rp = a(1-e) to get ~ 6543km... then the altitude zp = 6543- 6378 = 165km.. which is incorrect... can anyone help?? Thanks.

2. Jan 29, 2008

### Shooting Star

The eqn for r in a central force is given by:

½ mr’^2 + L^2/[2m(r^2)] + V = E,

where E is total energy, L is the ang mom, m is the mass of the body and V=PE. In this case, V= -GMm/r = -k/r.

In the 1st case, r=R is const. Because it’s a bound orbit, the total energy is, say, -E1, where E1 is +ve. Since r’=0, we can find the relationship between L, R etc from,

L^2/[2m(R^2)] – k/R = -E1. --(1)

Next, energy E is added to the body’s energy, but the L remains the same, since the impulse was in the direction of r. (E = ½ m*240^2 units.) The total energy is –E2. Then,

½ mr’^2 + L^2/[2m(r^2)] + V = -E2, where –E1+E= -E2, =>

½ mr’^2 = -E2 + k/r - L^2/[2m(r^2)] --(2).

At the apogee or the perigee, r’=0, from which you get the two reqd values of r. You can also show that r must lie between the two roots, showing that the orbit is bounded.

3. Jan 29, 2008

### GreenLRan

Thanks shooting star