Question

1. What is the potential energy of a 200kg satellite orbiting at 125 miles above the surface of the Earth?

a.) What is the satellite’s final velocity on impact, if it falls to the earth?

Answer 1

here,

mass of satellite , m1 = 200 kg

mass of Earth , m2 = 5.98 * 10^24 kg

radius of earth , r = 6.371 * 10^6 m

altitude , A = 125 miles

A = 2.01 * 10^5 m

the potential energy , PE = - G * m1 * m2 /(r + A)

PE = - 6.67 * 10^-11 * 200 * 5.98 * 10^24 /(6.371 * 10^6 + 2.01 * 10^5) J

PE - 1.21 * 10^10 J

a)

let the satellite's final velocity of impact be v

using conservation of energy

0.5 * m1 * v^2 = - G * m1 * m2 /(r) - ( -G * m1 * m2 /(r + A))

0.5 * v^2 = 6.67 * 10^-11 * 5.98 * 10^24 * ( 1/(6.371 * 10^6) - 1/(6.371 * 10^6 + 2.01 * 10^5))   

solving for v

v = 1957 m/s

the final velocity of impact is 1957 m/s