Question 299: Gravitation

Question

The height above the earth's surface at which the acceleration due to gravity becomes $\left(\frac{1}{n}\right)$ times the value at the surface is ( $R=$ radius of earth)

Accepted Answer

To find the height above the Earth's surface where the acceleration due to gravity ($g'$) becomes $\frac{1}{n}$ times the value at the surface ($g$), we can use the formula for gravitational acceleration at a distance:
$$ g' = \frac{GM}{(R+h)^2} $$
where:
$G$ is the gravitational constant,
$M$ is the mass of the Earth,
$R$ is the radius of the Earth,
$h$ is the height above the Earth's surface,
$g = \frac{GM}{R^2}$ is the acceleration due to gravity at the surface of the Earth.
Given that $g' = \frac{1}{n} \cdot g$, we have:
$$ \frac{GM}{(R+h)^2} = \frac{1}{n} \cdot \frac{GM}{R^2} $$
The gravitational constant $G$ and the Earth's mass $M$ cancel out, leading to:
$$ \frac{1}{(R+h)^2} = \frac{1}{n \cdot R^2} $$
Taking the reciprocal of both sides and then taking the square root:
$$ (R + h)^2 = n \cdot R^2 $$
$$ R + h = \sqrt{n} \cdot R $$
Isolating $h$, we find:
$$ h = \sqrt{n} \cdot R - R $$
Therefore, the height $h$ is:
$$ h = (\sqrt{n} - 1) \cdot R $$
The correct answer is Option D: $(\sqrt{n} - 1) \cdot R$.

Answer

$\frac{\mathrm{R}}{\sqrt{\mathrm{n}}}$

Answer

$\mathrm{R} \cdot \sqrt{\mathrm{n}}$

Answer

$\quad(\sqrt{n}+1) R$

Answer

$(\sqrt{\mathrm{n}}-1) \mathrm{R}$

Gravitation

The height above the earth's surface at which the acceleration due to gravity becomes \left(\frac{1}{n}\right) times the value at the surface is ( R= radius of earth)

Choose the correct answer: