Question

The acceleration due to gravity at a height of 6400 km from the surface of the earth is \(2.5 \, \text{ms}^{-2}\). The acceleration due to gravity at a height of 12800 km from the surface of the earth is (Radius of the earth = 6400 km) 

• A. \(1.11 \, \text{ms}^{-2}\)
• B. \(1.5 \, \text{ms}^{-2}\)
• C. \(2.22 \, \text{ms}^{-2}\)
• D. \(1.25 \, \text{ms}^{-2}\)

Hint:

The acceleration due to gravity decreases with height according to the inverse square law. When the height is comparable to the Earth's radius, the formula \( g_h = g_0 \left( \frac{R}{R+h} \right)^2 \) should be used instead of the simpler linear approximation.

Answer 1

The Correct Option is A

Step 1: Formula for Acceleration due to Gravity at a Height \( h \)
The acceleration due to gravity at a height \( h \) from the surface of the Earth is given by: \[ g_h = g_0 \left( \frac{R}{R+h} \right)^2 \] where: - \( g_h \) is the acceleration due to gravity at height \( h \), - \( g_0 \) is the acceleration due to gravity at the Earth's surface, - \( R \) is the radius of the Earth. Step 2: Given Values
We are given: \[ g_{6400} = 2.5 \, \text{ms}^{-2}, \quad R = 6400 \text{ km} \] Step 3: Ratio of Gravity at Different Heights
Using the formula: \[ \frac{g_{12800}}{g_{6400}} = \left( \frac{R+6400}{R+12800} \right)^2 \] \[ \frac{g_{12800}}{2.5} = \left( \frac{6400+6400}{6400+12800} \right)^2 \] \[ = \left( \frac{12800}{19200} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9} \] Step 4: Calculating \( g_{12800} \)
\[ g_{12800} = 2.5 \times \frac{4}{9} = \frac{10}{9} = 1.11 \, \text{ms}^{-2} \] Step 5: Conclusion
Thus, the acceleration due to gravity at a height of 12800 km is: \[ \boxed{1.11 \, \text{ms}^{-2}} \]