# Physics Gravitation Help?

A satellite is put into an elliptical orbit around the Earth. When the satellite is at its perigee, its nearest point to the Earth, its height above the ground is hp=213.0 km, and it is moving with a speed of vp=9.850 km/s. The gravitational constant G equals 6.67×10^−11 and the mass of Earth equals 5.972×10^24 kg. When the satellite reaches its apogee, at its farthest point from the Earth, what is its height ha above the ground? For this problem, choose the gravitational potential energy of the satellite to be 0 at an infinite distance from Earth.

### 1 Answer

- NCSLv 74 weeks agoFavorite Answer
For a particle, angular momentum = m*v*r

which is conserved since no external torques are applied.

So

m * vp * rp = m * va * ra

where rp = R + hp

and R is the Earth's radius. So after converting km to meters and canceling the mass,

9850m/s * (6371+213.0)x10³m = va * (6371+ha)x10³m

6.4852x10^7 = va * (6371+ha)

for va in m/s and ha in km

so

va = 6.4852x10^7 / (6371+ha)

Next we can conserve total mechanical energy:

½m*vp² - GmM/rp = ½m*va² - GmM/ra

mass cancels, multiply by 2

vp² - 2GM/(6371+213.0)x10³ = va² - 2GM/(6371+ha)x10³

plug in values and substitute for va (and drop units for ease -- ha is in kilometers)

9850² - 2*6.674x10^(−11)*5.972x10^24/6584x10³ =

(6.4852x10^7 / (6371+ha))² - 2*6.674x10^(−11)*5.972x10^24/((6371+ha)x10³)

This is quadratic with solutions at

ha = 212.9 km

and ha = 20 190 km ◄

The first solution is the perigee point (or close enough -- 213 km). The second is the apogee point.

Whew!

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