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

To determine the acceleration due to gravity at a height of 12800 km from the surface of the earth, we use the formula for gravitational acceleration at a height \( h \):

\( g_h = g_0 \left(\frac{R}{R+h}\right)^2 \)

where:

• \( g_h \) is the acceleration due to gravity at height \( h \).
• \( g_0 \) is the acceleration due to gravity on the surface.
• \( R \) is the radius of the earth.

Given:

• Height from the surface: 6400 km, \( g_h = 2.5 \, \text{ms}^{-2} \).
• Radius of the earth \( R = 6400 \, \text{km} \).

First, calculate the acceleration due to gravity at a height of 6400 km:

\( g_1 = g_0 \left(\frac{R}{2R}\right)^2 = g_0 \left(\frac{1}{2}\right)^2 = \frac{g_0}{4} \)

Since \( g_1 = 2.5 \, \text{ms}^{-2} \), we have \( g_0/4 = 2.5 \), thus:

\( g_0 = 10 \, \text{ms}^{-2} \)

Next, calculate the acceleration due to gravity at a height of 12800 km:

\( g_2 = g_0 \left(\frac{R}{R+h}\right)^2 \)

Here, \( h = 12800 \, \text{km} = 2R \):(since \( h = 12800 \, \text{km} \), add to \( R \) to find the distance from the center as \( R + 2R = 3R \))

\( g_2 = 10 \left(\frac{6400}{19200}\right)^2 = 10 \left(\frac{1}{3}\right)^2 = 10 \times \frac{1}{9} = 1.11 \, \text{ms}^{-2} \)

Thus, the acceleration due to gravity at a height of 12800 km from the surface of the earth is \(\boxed{1.11 \, \text{ms}^{-2}}\).

Answer 2

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: \[ \boxed{1.11 \, \text{ms}^{-2}} \]