# Calculating satellite's angular speed at perihelion

## Homework Statement

A satellite is in elliptical orbit with a period of 8.00 x 10^4 s about a planet of mass 7.00 x 10^24 kg. At aphelion, at radius 4.5 x 10^7 m, the satellite's angular speed is 7.158 x 10^-5 rad/s. What is its angular speed at perihelion?

## Homework Equations

V(orbital) = sqrt(GM/R)

T=2∏sqrt(R^3/GM)

V=ωr

## The Attempt at a Solution

I used Kepler's 3rd law to find the semimajor axis of the satellite:

T=2∏sqrt(R^3/GM)
8.00 x 10^4 s = 2∏sqrt(R^3/(6.67 x 10^.11)(7 x 10^24 kg))

and found that the semimajor axis (R) is 4.23 x 10^7 m

I then reasoned that twice the semimajor axis minus the aphelion distance = the perihelion distance, so

2(4.23 x 10^7 m) - 4.5 x 10^7 m = perihelion distance

Perihelion distance (r) = 3.96 x 10^7 m

I then used the value of the radius at perihelion (r) in the V(orbital) = sqrt(GM/r) expression and found:

V(orbital) = sqrt[((6.67 x 10^-11)(7 x 10^24 kg))/(3.96 x 10^7 m)]
V(orbital) = 3433.65 m/s

I then plugged this into V=ωr:

3433.65 m/s = ω x 3.96 x 10^7 m
ω = 8.67 x 10^-5 rad/s

However, the answer should be 9.29 x 10^-5 rad/s. My answer key shows that the conservation of angular momentum was used to find the answer. However, I don't see why my approach doesn't also work. It seems to yield a similar answer, but just far off enough to make me think that the approach is flawed. Any help would be greatly appreciated!

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ehild
Homework Helper
The equation you used V(orbital) = sqrt(GM/R) is valid for R as the radius of curvature, which is equal to the radius of orbit only for circular motion.

Use conservation of momentum.

ehild

Ok, will do. Thanks!