Question

The height above the surface of Earth at which gravitational field intensity is reduced to $1\%$ of its value on the surface of Earth is ($R_e$ is radius of Earth)

• A. $100R_e$
• B. $10R_e$
• C. $99R_e$
• D. $9R_e$

Answer 1

The correct option is D $9R_e$

Gravitational field at the surface of the Earth is given by,
$$E_s=\frac{GM}{R_e^2}...\left(1\right)$$ For a large spherical object like a planet, the gravitational field at a point outside it, at distance $h$ from the surface can be calculated by assuming the entire mass to be concentrated at its center,$$E_h=\frac{GM}{(R_e+h{)}^2}...\left(2\right)$$ According to the problem,

$E_h=1\%of{E}_g$

From equation $\left(1\right)$ and $\left(2\right)$, we get

$\frac{GM}{(R_e+h{)}^2}=0.01\times \frac{GM}{R_e^2}$

$\Rightarrow \frac{1}{(R_e+h{)}^2}=0.01\times \frac{1}{R_e^2}$

$\Rightarrow \frac{1}{(R_e+h{)}^2}=\frac{1}{100R_e^2}$

$\Rightarrow 10R_e=R_e+h$

$\therefore h=9R_e$

Hence, option (d) is the correct answer.

$\text{Key Concept:}$ For a large spherical object like a planet, the gravitational field at a point outside can be calculated by assuming the entire mass to be concentrated at its center. $\text{Why this question:}$ To make students familiarize with the formula for gravitational field intensity due to a point mass.