- #1

yasar1967

- 73

- 0

**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.

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

Then I put the value into:

1/2 mv^2 - GmM/R1 = 1/2mV^2

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?

[/b]