Question

Two satellites of same mass are launched in circular orbits at heights '$R$' and '$2R$' above the surface of the earth. The ratio of their kinetic energies is ($R = \text{radius of the earth}$)

• A. $1 : 3$
• B. $3 : 2$
• C. $4 : 9$
• D. $9 : 4$

Hint:

Always remember that orbital mechanics calculations require the distance measured from the center of the planet ($R+h$), not just the altitude above the surface! Converting heights of $1R$ and $2R$ to center distances of $2R$ and $3R$ lets you invert the values to find the kinetic energy ratio $\frac{3}{2}$ instantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Two identical satellites orbit at heights h₁ = R and h₂ = 2R above Earth's surface; find the ratio of their kinetic energies K₁ : K₂.

Step 2: Key Formula or Approach:
For a satellite, K = GMm/(2r) where r = R + h. Thus K ∝ 1/r.

Step 3: Detailed Explanation:
r₁ = R + R = 2R, r₂ = R + 2R = 3R. K₁/K₂ = r₂/r₁ = 3R/2R = 3/2, so ratio is 3 : 2.

Step 4: Final Answer:
Ratio of kinetic energies is 3 : 2, matching option (B).