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 = 229.0 km, and it is moving with a speed of up = 8.050 km/s. The gravitational constant G equals 6.67 x 10-11 mº.kg-1.5-2 and the mass of Earth equals 5.972 x 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. km

Answer 1

The height above the ground is 1279.51 km

Solution:

As per the question:

Height above the ground,
`h_(p) = 229.0\ km`

Speed of the satellite,
`v_(p) = 8.050\ km/s = 8050\ m/s`

Gravitational constant,
`G = 6.67* 10^(- 11)\ m.kg`

Mass of the Earth, m =
`5.972* 10^(24)\ kg`

Now,

The distance of the earth from the perigee is given by:

`R_(p) = h_(p) + R_(E)`

where

`R_(E) = 6371\ km`

`R_(p) = 229.0 + 6371 = 6600\ km`

Now,

To calculate the distance of the earth from the earth from the apogee:

`R_(A) = (R_(p))/((2Gm)/(v_(p)^(2)R_(p) - 1))`

`R_(A) = (6600* 1000)/((2* 6.67* 10^(- 11)* 5.972* 10^(24))/(8050^(2)6600* 1000 ) - 1)}`

`R_(A) = 7650.51\ km`

Height above the ground,
`h_(A) = R_(A) - R_(E) = 7650.51 - 6371 = 1279.51\ km`