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)

Answer 2

Step 1: Understand the problem.
We are asked to find the correct formula for the height of a satellite from Earth's surface.
The height of a satellite is related to the orbital period \( T \), the radius of Earth \( R \), and the gravitational constant \( g \). The formula for the height of the satellite is based on the relationship between these quantities.

Step 2: Derivation of the formula.
The gravitational force provides the centripetal force that keeps the satellite in orbit. The centripetal force is given by: \[ \frac{mv^2}{r} = \frac{GMm}{r^2}, \] where \( m \) is the mass of the satellite, \( v \) is its velocity, \( r \) is the distance from the center of the Earth to the satellite, and \( G \) is the gravitational constant. For a satellite in a circular orbit, we can relate its velocity \( v \) to its orbital period \( T \) by the equation: \[ v = \frac{2\pi r}{T}. \] Substituting this into the centripetal force equation, we get: \[ \frac{m \left( \frac{2\pi r}{T} \right)^2}{r} = \frac{GMm}{r^2}. \] Simplifying this expression gives: \[ \frac{4\pi^2 r}{T^2} = \frac{GM}{r^2}. \] Solving for \( r \), we get: \[ r^3 = \frac{4\pi^2 GM}{T^2}. \] The gravitational constant \( G \) is related to the gravitational acceleration \( g \) at the surface of the Earth by \( g = \frac{GM}{R^2} \). So we can substitute \( GM = gR^2 \) into the equation: \[ r^3 = \frac{4\pi^2 gR^2}{T^2}. \] Taking the cube root of both sides: \[ r = \left( \frac{T^2 R^2 g}{4 \pi^2} \right)^{1/3}. \] Finally, the height \( h \) of the satellite from the surface of the Earth is given by the distance from the Earth's center minus the radius of the Earth: \[ h = r - R = \left( \frac{T^2 R^2 g}{4 \pi^2} \right)^{1/3} - R. \]

Step 3: Conclusion.
Thus, the correct formula for the height of the satellite from Earth's surface is:
\[ \boxed{ \left( \frac{T^2 R^2 g}{4 \pi^2} \right)^{1/3} - R }. \]