Question

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=215.0 km, and it is moving with a speed of ?p=8.850 km/s. The gravitational constant ? equals 6.67×10−11 m3·kg−1·s−2 and the mass of Earth equals 5.972×1024 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 gravitational potential energy of the satellite to be 0 at an infinite distance from Earth.

Answer 1

m = mass of satellite

M = mass of earth = 5.972 x 1024 kg

R = radius of earth = 6.371 x 106 m

hp = height above the surafe at perigee = 215 km = 0.215 x 106 m

vp = speed at perigee = 8850 m/s

ha = height above the surafe at apogee = ?

va = speed at apogee = ?

Using conservation of angular momentum

m vp (R + hp ) = m va (R + ha)

vp (R + hp ) = va (R + ha)

va = vp (R + hp ) /(R + ha) Eq-1

Using conservation of energy

- GMm/(R + hp ) + (0.5) m vp2 = - GMm/(R + ha ) + (0.5) m va2

- GM/(R + hp ) + (0.5) vp2 = - GM/(R + ha ) + (0.5) va2

Using Eq-1

- GM/(R + hp ) + (0.5) vp2 = - GM/(R + ha ) + (0.5) (vp (R + hp ) /(R + ha))2

- (6.67 x 10-11)(5.972 x 1024)/((6.371 x 106) + (0.215 x 106) ) + (0.5) (8850)2 = - (6.67 x 10-11)(5.972 x 1024)/((6.371 x 106) + ha ) + (0.5) ((8850)((6.371 x 106) + (0.215 x 106) ) /((6.371 x 106) + ha))2

- 2.13 x 107 = - (6.67 x 10-11)(5.972 x 1024)/((6.371 x 106) + ha ) + (0.5) ((8850)((6.371 x 106) + (0.215 x 106) ) /((6.371 x 106) + ha))2

ha = 5.75 x 106 m