# Homework Help: Satellite Problem

1. Feb 24, 2008

### yasar1967

A small package is fired off Earth's surface with a speed V at a 45o angle. It reaches maximum height of h above surface at h=6370km, a value equal to Earth's radius itself.
What is the speed when it reaches this height? Ignore any effects that might come from Earth's rotation.

Energy converstion: 1/2 mv^2 - GmM/R1 = 1/2mV^2 - 1/2 GmM/R2
Parabolic Motion: hmax= vo^2(sin^2 ø) /2g
Angular momentum conversition :L=rmv etc. etc.

3. I first calculated the Vo speed from hmax= vo^2(sin^2 ø) /2g equation
Then I put the value into:
1/2 mv^2 - GmM/R1 = 1/2mV^2
- 1/2 GmM/R2

noting that R2=hmax=2 x Rearth

Then I calculated and found the SAME result from parabolic motion equations:
first
time= x/V0 cosø ... and then finding out Vx and Vy thus resulting to V speed.

Yet,

The book starts with angular momentum conversion, stating that :
m(Vcosø)R = m Vfinal 2R and then using Vescape speed formula and energy conversition it comes out with completely different result.

Where did I go wrong? or how come results do not reconcile?
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1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 24, 2008

### kamerling

hmax = vo^2(sin^2 ø) /2g is only valid if the acceleration of gravity is constant.

You can use the angular momentum equation to get the velocity at the apogee of the orbit as a function of the velocity at the surface. You can then subtitute that in the energy equation