Question

The value of g at the earth's surface is 10 $m{s}^{-2}$, the value of g at a height of 3 $R_E$ from the surface of earth is

• A. $\frac{10}{9}m{s}^{-2}$
• B. $\frac{5}{8}m{s}^{-2}$
• C. $\frac{3}{8}m{s}^{-2}$
• D. $\frac{10}{8}m{s}^{-2}$

Answer 1

The correct option is B

$\frac{5}{8}m{s}^{-2}$

Acceleration due to gravity:

1. The acceleration achieved by any object due to the gravitational force of attraction by any planet is called acceleration due to gravity by the earth.
2. Due to the fact that each planet has a varied mass and radius, the acceleration due to gravity will vary.

Given:

g at the surface of the earth is $10m/s^2$

At a height of h, acceleration due to gravity can be calculated using $g\text{'}=\frac{g}{{(1+{\displaystyle \frac{h}{R_E}})}^2}$ where $R_E$ is the radius of earth.

At height h=$R_E$, acceleration due to gravity ($g^{\text{'}}$) = $=\frac{g}{{(1+{\displaystyle \frac{3R_E}{R_E}})}^2}$

$g^{\text{'}}=\frac{g}{{(1+3)}^2}=\frac{10}{16}=\frac{5}{8}m/s^2$

Therefore the right answer is an option (b) $\frac{5}{8}m/s^2$