Question

A 1200 kg artificial satellite is in an orbit of radius $2R_{E}$ about the earth. The energy required to transfer it to an orbit of radius $3R_{E}$ is ($g=10\text{ms}^{-2}$, $R_{E}=6400 \text{ km}$)

• A. $1.2\times10^{9} \text{ J}$
• B. $6.4\times10^{9} \text{ J}$
• C. $6400\times10^{3} \text{ J}$
• D. $3.2\times10^{9} \text{ J}$

Hint:

Remember that orbital shift energy $\Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$. Substitute $GM = gR_E^2$ for easy numerical calculations.

Answer 1

The Correct Option is B

Step 1: Total energy of a satellite.
A satellite in a circular orbit of radius $r$ has total energy $E = -\dfrac{GMm}{2r}$. The minus sign shows it is bound.

Step 2: Replace $GM$ with $gR_{E}^{2}$.
Since $g = \dfrac{GM}{R_{E}^{2}}$, we can write $GM = gR_{E}^{2}$, so $E = -\dfrac{mgR_{E}^{2}}{2r}$.

Step 3: Energy in the first orbit ($r=2R_{E}$).
\[ E_{1} = -\frac{mgR_{E}^{2}}{2(2R_{E})} = -\frac{mgR_{E}}{4} \]
Step 4: Energy in the second orbit ($r=3R_{E}$).
\[ E_{2} = -\frac{mgR_{E}^{2}}{2(3R_{E})} = -\frac{mgR_{E}}{6} \]
Step 5: Energy needed is the difference.
\[ \Delta E = E_{2} - E_{1} = mgR_{E}\left(\frac{1}{4} - \frac{1}{6}\right) = \frac{mgR_{E}}{12} \]
Step 6: Put in the numbers.
With $m=1200$, $g=10$, $R_{E}=6.4\times 10^{6}$ m: \[ \Delta E = \frac{1200\times 10\times 6.4\times 10^{6}}{12} = 6.4\times 10^{9} \text{ J} \]This is option 2.
\[ \boxed{6.4\times 10^{9} \text{ J}} \]