Question

Correct formula for height of a satellite from earths surface is :

• A. \( \left( \frac{T^2 R^2 g}{4 \pi} \right)^{1/2} - R \)
• B. \( \left( \frac{T^2 R^2 g}{4 \pi^2} \right)^{1/3} - R \)
• C. \( \left( \frac{T^2 R^2}{4 \pi^2 g} \right)^{1/3} - R \)
• D. \( \left( \frac{T^2 R^2}{4 \pi^2 g} \right)^{-1/3} + R \)

Answer 1

The Correct Option is B

1. **Using Gravitational Force and Centripetal Force:**
For a satellite of mass m orbiting Earth at height h above the surface, the gravitational force provides the required centripetal force:
GMm / (R + h)² = mv² / (R + h).
Simplifying, we get:
GM / (R + h) = v². (1)

2. **Relate Orbital Velocity to Period:**
The orbital velocity v can also be expressed in terms of the orbital period T:
v = 2π(R + h) / T. (2)

3. **Equate Gravitational Force with Centripetal Acceleration:**
We know that GM = gR² (where g is the acceleration due to gravity on Earth's surface). Substituting this in equation (1):
gR² / (R + h) = v².

4. **Combine Equations (1) and (2):**
Substitute v from equation (2) into the above expression:
gR² / (R + h) = (2π(R + h) / T)².
Rearranging gives:
T²R²g / (2π)² = (R + h)³.

5. **Solve for Height h:**
Taking the cube root of both sides, we get:
R + h = (T²R²g / 4π²)^(1/3).
Therefore,
h = (T²R²g / 4π²)^(1/3) - R.

**Answer:** (T²R²g / 4π²)^(1/3) - R (Option 2)