Question

The value of acceleration due to gravity at a height of \( 4R_E \) from surface of earth is \[ R_E \text{ is radius of earth and acceleration due to gravity on the surface of the earth = 10 ms}^{-2} \]

• A. \( 0.2 \, \text{ms}^{-2} \)
• B. \( 1 \, \text{ms}^{-2} \)
• C. \( 0.4 \, \text{ms}^{-2} \)
• D. \( 3 \, \text{ms}^{-2} \)

Hint:

For integrals involving trigonometric functions like \( \sin \) and \( \cos \), use the identity \( \sin^2 x + \cos^2 x = 1 \) to simplify the expression and check if symmetry or known integrals can help to evaluate the result.

Answer 1

The Correct Option is C

The acceleration due to gravity \( g_h \) at height \( h \) from the surface of the earth is given by the formula: \[ g_h = \frac{g}{(1 + \frac{h}{R_E})^2} \] where \( g \) is the acceleration due to gravity on the surface of the earth and \( R_E \) is the radius of the earth. Here, \( h = 4R_E \) (height above the earth's surface), so the formula becomes: \[ g_h = \frac{g}{(1 + \frac{4R_E}{R_E})^2} = \frac{10}{(1 + 4)^2} = \frac{10}{25} = 0.4 \, \text{ms}^{-2} \] Hence, the value of the acceleration due to gravity at a height of \( 4R_E \) from the surface of the earth is \( 0.4 \, \text{ms}^{-2} \).

Answer 2

Step 1: Use the formula for gravity at height above Earth's surface.
The acceleration due to gravity at a height \( h \) above the Earth's surface is given by: \[ g_h = g \left( \frac{R_E}{R_E + h} \right)^2 \] Where:
- \( g = 10 \, \text{ms}^{-2} \) (on Earth's surface)
- \( h = 4R_E \) (height given is 4 times the radius of Earth)
- So, total distance from Earth’s center = \( R_E + 4R_E = 5R_E \)

Step 2: Plug into the formula.
\[ g_h = 10 \left( \frac{R_E}{5R_E} \right)^2 = 10 \left( \frac{1}{5} \right)^2 = 10 \times \frac{1}{25} = 0.4 \, \text{ms}^{-2} \]

Final Answer: \( \boxed{0.4 \, \text{ms}^{-2}} \)