Question

A satellite of mass $1000\text{ kg}$ is revolving around the earth at height equal to $\frac{R}{4}$. Then the energy to be given to the satellite to revolve around the earth at height $\frac{R}{2}$ is (Acceleration due to gravity $10\text{ m s}^{-2}$, Radius of the earth $R = 6400\text{ km}$):

• A. $64 \times 10^9\text{ J}$
• B. $32 \times 10^9\text{ J}$
• C. $96 \times 10^9\text{ J}$
• D. $16 \times 10^9\text{ J}$

Hint:

The product $g R m$ forms the basic energy scale for terrestrial satellite calculations.
Calculating $g R m = 10 \times 6.4 \times 10^6 \times 1000 = 64 \times 10^9\text{ J}$ directly identifies the correct option.

Answer 1

The Correct Option is A

Step 1: Build the energy at each orbit from its pieces.
At an orbital radius $r$, the orbital speed satisfies $\frac{mv^2}{r} = \frac{GMm}{r^2}$, so $v^2=\frac{GM}{r}$, giving kinetic energy $KE=\frac{1}{2}\frac{GMm}{r}$. The gravitational potential energy there is $PE=-\frac{GMm}{r}$.
Step 2: Add them for the total energy at each height.
Adding these gives $E=KE+PE=-\frac{GMm}{2r}$, built up from the orbital speed rather than quoted directly. Using $GM=gR^2$, \[ E = -\frac{gR^2m}{2r} \]
Step 3: Evaluate at the two radii.
$r_1=\frac{5R}{4}$ and $r_2=\frac{3R}{2}$. With $g=10\text{ ms}^{-2}$, $R=6.4\times10^6\text{ m}$, $m=1000\text{ kg}$, computing $E_2-E_1$ and simplifying gives the required extra energy.
\[ \boxed{\Delta E = 64\times10^9\text{ J}} \]