If the value of the gravitational acceleration on the surface of the Earth is [tex]g[/tex], then what will be its value at a distance of [tex]h = 6.38 \times 10^6 \, \text{m}[/tex] from the surface of the Earth?

A. [tex]\frac{g}{8}[/tex]

B. [tex]\frac{g}{4}[/tex]

C. [tex]\frac{g}{2}[/tex]

D. [tex]g[/tex]



Answer :

To solve the problem of finding the gravitational acceleration at a height [tex]\( h \)[/tex] from the surface of the Earth, let's go through the relevant formulas and applications step-by-step.

1. Gravitational Acceleration on Earth's Surface:
The gravitational acceleration [tex]\( g \)[/tex] on the Earth's surface is approximately [tex]\( 9.8 \ \text{m/s}^2 \)[/tex].

2. Distance from the Center of the Earth:
Given:
- Radius of the Earth [tex]\( R = 6.38 \times 10^6 \ \text{m} \)[/tex]
- Height above the surface [tex]\( h = 6.38 \times 10^6 \ \text{m} \)[/tex]

So, the total distance from the center of the Earth is [tex]\( R + h = 6.38 \times 10^6 + 6.38 \times 10^6 = 2 \times 6.38 \times 10^6 \ \text{m} \)[/tex].

3. Gravitational Acceleration at Height [tex]\( h \)[/tex]:
The formula for gravitational acceleration at a height [tex]\( h \)[/tex] from the surface of the Earth is given by:
[tex]\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \][/tex]

4. Simplifying the Expression:
Substituting [tex]\( h = R \)[/tex] into the formula:
[tex]\[ g' = \frac{g}{(1 + \frac{R}{R})^2} = \frac{g}{(1 + 1)^2} = \frac{g}{2^2} = \frac{g}{4} \][/tex]

Therefore, the gravitational acceleration at a distance of [tex]\( h = 6.38 \times 10^6 \ \text{m} \)[/tex] from the surface of the Earth is:

[tex]\(\boxed{\frac{g}{4}}\)[/tex]

Given the multiple-choice options, the correct answer is B. [tex]\(\frac{g}{4}\)[/tex]. This simplifies to the numerical value 2.45 m/s² when using [tex]\( g = 9.8 \)[/tex] m/s².