Question

The ratio of energy required to raise a satellite of mass $m$ to height $h$ above the earth's surface to that required to put it into the orbit at same height is ($R$ = radius of earth)

• A. $\frac{h}{R}$
• B. $\frac{2h}{R}$
• C. $\frac{3h}{R}$
• D. $\frac{h}{2R}$

Hint:

For small orbital altitudes close to Earth ($h \ll R$), you can approximate the potential energy change using the standard formula $W_1 \approx mgh$.
The kinetic energy required to orbit at the surface is $E_k = \frac{1}{2}mv^2 = \frac{1}{2}m(\sqrt{gR})^2 = \frac{1}{2}mgR$.
Taking the ratio gives $\frac{mgh}{\frac{1}{2}mgR} = \frac{2h}{R}$, which serves as a quick way to verify the expression!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem breaks down a two-stage satellite launch sequence: 1.

Stage 1: Raising the satellite of mass $m$ vertically from the Earth's surface up to a target altitude $h$ without circular speed. 2.

Stage 2: Accelerating the satellite at that altitude to its critical orbital velocity so it enters a stable circular orbit.
We need to find the ratio of the energy expended in Stage 1 ($W_1$) to the kinetic energy required in Stage 2 ($E_k$).

Step 2: Key Formula or Approach:
3. The energy required to lift a mass to a height $h$ is equal to the change in its gravitational potential energy: $$W_1 = \Delta U = U_{\text{at } h} - U_{\text{surface}} = -\frac{GMm}{R+h} - \left(-\frac{GMm}{R}\right) = \frac{GMmh}{R(R+h)}$$ 4. The kinetic energy required to keep a satellite in a circular orbit at that altitude depends on its orbital velocity $v_c = \sqrt{\frac{GM}{R+h}}$: $$E_k = \frac{1}{2}mv_c^2 = \frac{GMm}{2(R+h)}$$

Step 3: Detailed Explanation:
Let's find the ratio of the energy required to raise the satellite ($W_1$) to the energy required to orbit it ($E_k$): $$\text{Ratio} = \frac{W_1}{E_k} = \frac{\frac{GMmh}{R(R+h)}}{\frac{GMm}{2(R+h)}}$$ Cancel out the common factors $GMm$ and $(R+h)$ from both the numerator and denominator: $$\text{Ratio} = \frac{\frac{h}{R}}{\frac{1}{2}} = \frac{2h}{R}$$ This matches the expression in option (B).

Step 4: Final Answer:
The ratio of the energy required is $\frac{2h}{R}$, which corresponds to option (B).