Question

The $75,000kg$space shuttle used to fly in a $250km$high circular orbit. It needed to reach a$610km$ high circular orbit to service the Hubble Space Telescope. How much energy was required to boost it to the new orbit?

Short answer

$1.16\times {10}^{11}J$excess energy is required to move the satellite from$250km$to$610km$.

Step-by-step solution

Given Information

The satellite has mass, $m=75000kg$.

Initial radius of the circular orbit is $250km=250000m$.

New radius of the circular orbit is $610km=610000m$.

Formula used

Excess Energy, $∆E=\frac{GMm}{2R}-\frac{GMm}{2\left(R+r\right)}----\left(i\right)$.

Here, gravitational constant $G=6.67\times {10}^{-11}N·m^2/k{g}^2$.

Mass of the space shuttle, localid="1648190643516" $M=75000kg$.

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

Radius of the orbit of space shuttle from earth's centre = Radius of earth + Distance of the space shuttle from earth's surface$=\left(6.384\times {10}^6+250000\right)m$.

New Radius of the orbit of space shuttle from earth's centre localid="1648189785883" $R+r=$Radius of earth $+$New distance of the space shuttle from earth's surface localid="1648190353993" $=\left(6.384\times {10}^6+610000\right)m$.

Calculation

Substitute the values in equation $\left(i\right)$to obtain $∆E$.

role="math" localid="1648190453466" $$\begin{gathered} ∆E=\frac{GMm}{2R}-\frac{GMm}{2\left(R+r\right)} \\ ∆E=\frac{GMm}{2}\left[\frac{1}{R}-\frac{1}{R+r}\right] \\ =\frac{6.67\times {10}^{-11}N·m^2/k{g}^2\times 75000kg\times 5.972\times {10}^{24}kg}{2}\left[\frac{1}{\left(6.384\times {10}^6+250000\right)m}-\frac{1}{\left(6.384\times {10}^6+610000\right)m}\right] \\ =1.16\times {10}^{11}J \end{gathered}$$

Final Answer

$1.16\times {10}^{11}J$excess energy is required to move the satellite from $250km$ to $610km$.