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 are orbiting the Earth at different altitudes. The first satellite is at a height $h_1 = R$ and the second is at a height $h_2 = 2R$ above the surface. We need to determine the ratio of their kinetic energies, $K_1 : K_2$.

Step 2: Key Formula or Approach:
The kinetic energy of a satellite of mass $m$ in a stable circular orbit at a distance $r$ from the center of a planet of mass $M$ is given by: $$K = \frac{GMm}{2r}$$ The total orbital radius from the center of the Earth is $r = R + h$, where $R$ is the Earth's radius and $h$ is the altitude above the surface. Therefore, kinetic energy is inversely proportional to the orbital radius: $K \propto \frac{1}{r}$.

Step 3: Detailed Explanation:
Let's find the total orbital radius for each satellite from the center of the Earth: $$\text{For Satellite 1: } r_1 = R + h_1 = R + R = 2R$$ $$\text{For Satellite 2: } r_2 = R + h_2 = R + 2R = 3R$$ Since $K \propto \frac{1}{r}$, we set up the ratio of their kinetic energies: $$\frac{K_1}{K_2} = \frac{r_2}{r_1}$$ Substitute our values for $r_1$ and $r_2$: $$\frac{K_1}{K_2} = \frac{3R}{2R} = \frac{3}{2}$$ This corresponds to a ratio of $3 : 2$.

Step 4: Final Answer:
The ratio of their kinetic energies is $3 : 2$, which corresponds to option (B).