Question

A spacecraft of 110 kg mass is in a circular orbit about the Earth at a height h = 2RE.

(a) What is the period of the spacecraft's orbit about the Earth? T = . h

(b) What is the spacecraft's kinetic energy? K = . J

(c) Express the angular momentum L of the spacecraft about the center of the Earth in terms of its kinetic energy K. (Use the following as necessary: RE for the radius of the Earth, K for the kinetic energy of the satellite, and m for the mass of the satellite.) L =

(d) Find the numerical value of the angular momentum. L = J · s

Answer 1

Given

Mass of the space craft m = 110 Kg

Height h = 2R

Mass of the earth M = 5.98 x 10²⁴ kg

Radius of the earth R = 6.38 x 10⁶ m

Gravitational constant G = 6.67 x 10^-11 Nm²/kg²

Acceleration due to gravity g = 9.8 m/s²

Solution

a)

Time period T = 2π√{(R + h)³/GM}

= 2 x 3.14 x √(9R³/GM)

= 5067.30 s

= 1.41 hours

b)

kinetic energy K = GMm/2(R+h)

= 4387.526 x 10¹³ / 6R

= 114.62 x 10⁷ J

c)

L = mv(R+h)

= 3mvR

= 3R * 2K /v

= 6KR/v

V = √(GM/3R)

So

L = 6KR *√(3R/GM)

d)

L = 6 x 114.62 x 10⁷ x 6.38 x 10⁶ √(3x6.38 x 10⁶ / 6.67 x 10^-11 x 5.98 x 10²⁴)

= 9611.47 x 10⁹

= 9.6 x 10¹² Js