Question

The 65,000 kgkg space shuttle used to fly in a 250-km-high circular orbit. It needed to reach a 610-km-high circular orbit to service the Hubble Space Telescope.
How much energy was required to boost it to the new orbit?

Answer 1

To solve this problem we will apply the concepts related to gravitational potential energy. The existing difference between the two heights will allow us to find the net energy to perform the action determined in the statement. The energy required to boost the shuttle to the new orbit is

$\Delta E = E_2-E_1$

$\Delta E = GMm(\frac{1}{2r_1}-\frac{1}{2r_2})$

Our values are

$M = 5.972*10^{24}kg \rightarrow$ Mass of Earth

$R = 6.371*10^6m \rightarrow$ Radius of Earth

$G = 6.674*10^{-11}m^3/kgs^2 \rightarrow$ Gravitational Universal Constant

$m = 65000kg$

The initial radius of the orbit of the shuttle is

$r_1 = R+h_1$

$r_1 = 6.371*10^6+2.5*10^5$

$r_1 = 6.621*10^6m$

Final Orbit radius of the shuttle is

$r_2 = R+h_2$

$r_2 = 6.371*10^6+6.1*10^5$

$r_2 = 6.98*10^6m$

Replacing we have that

$\Delta E = (6.674*10^{-11})(5.972*10^{24})(65000)(\frac{1}{2(6.621*10^6)}-\frac{1}{2(6.98*10^6)})$

$\Delta E = 1.04*10^{11}J$