Question

A 3500 kg spaceship is in a circular orbit 220 km above the surface of Earth. It needs to be moved into a higher orbit of 360 km to link up with the space station at that altitude. In this problem you can take the mass of the Earth to be (5.97 * 10^24)a) How much work, in joules, do the spaceship’s engines have to perform to move to the higher orbit? Ignore any change of mass due to fuel consumption?

Answer 1

Answer:

The work done is calculated as $0.044\times 10^{11}\ J$

Solution:

As per the question:

Mass of the spaceship, m = 3500 kg

Height of the orbit, h = 220 km

Mass of the earth, $M_{e} = 5.97\times 10^{24}\ kg$

Height of the higher orbit, h' = 360 km

Radius of the orbit, $R = 6.37\times 10^{6}\ km$

Now,

The work done is given by the change in the gravitational potential energy:

Gravitational Potential Energy, U = $- \frac{GM_{e}m}{R + h}$

Now, for the lower orbit:

$U_{l} = - \frac{GM_{e}m}{R + h}$

$U_{l} = - \frac{6.67\times 10^{- 11}\times 5.97\times 10^{24}\times 3500}{6.37\times 10^{6} + 220\times 10^{3}}$

$U_{l} = - 2.115\times 10^{11}\ J$

For upper orbit:

$U_{U} = - \frac{GM_{e}m}{R + h}$

$U_{u} = - \frac{6.67\times 10^{- 11}\times 5.97\times 10^{24}\times 3500}{6.37\times 10^{6} + 360\times 10^{3}}$

$U_{U} = - 2.071\times 10^{11}\ J$

Change in the gravitational Potential energy:

$\Delta U = U_{U} - U_{l} = - 2.071\times 10^{11} - (- 2.115\times 10^{11}) = 0.044\times 10^{11}\ J$

Therefore, the work done:

$W = 0.044\times 10^{11}\ J$