(A) The gravitational force at the surface of the Earth is $F_s = \frac{GM_e m}{R_e^2}$.The gravitational force at height $h$ is $F_h = \frac{GM_e m}{

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) The gravitational force at the surface of the Earth is $F_s = \frac{GM_e m}{R_e^2}$.The gravitational force at height $h$ is $F_h = \frac{GM_e m}{(R_e + h)^2}$.According to the problem,$F_h = \frac{1}{3} F_s$.Substituting the expressions: $\frac{GM_e m}{(R_e + h)^2} = \frac{1}{3} \frac{GM_e m}{R_e^2}$.Simplifying the equation: $(R_e + h)^2 = 3 R_e^2$.Taking the square root on both sides: $R_e + h = \sqrt{3} R_e$.Therefore,$h = \sqrt{3} R_e - R_e = R_e(\sqrt{3} - 1)$.

Class 11 Physics Gravitation Acceleration Due to Gravity and its Variation

Medium

The gravitational force of the Earth on a rocket at a height $h$ from the surface of the Earth is $\frac{1}{3}$ of the gravitational force on the surface of the Earth. Find the relationship between $h$ and $R_e$ (the radius of the Earth).

A
$h = R_e(\sqrt{3} - 1)$
B
$h = R_e(\sqrt{3} + 1)$
C
$h = R_e(3 - 1)$
D
$h = R_e(\sqrt{3})$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

Gravitation

Subtopic

Acceleration Due to Gravity and its Variation

Derive the equation for the variation of $g$ due to height from the surface of the Earth.

Medium

View Solution

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (where $x$ and $h$ are much smaller than the radius of the earth $R_e$):

Medium

View Solution

The acceleration due to gravity on the Moon is $\frac{1}{6}$ times that of the Earth. If the ratio of the density of the Earth $(\rho_e)$ to the Moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,what is the radius of the Moon $(R_m)$ in terms of the radius of the Earth $(R_e)$?

Difficult

View Solution

The acceleration due to gravity on the surface of the Earth is $g$. What is the acceleration due to gravity at a height $h = R$ above the Earth's surface,where $R$ is the radius of the Earth?

Medium

View Solution

$A$ body weighs $200 \; N$ on the surface of the earth. How much will it weigh halfway down to the centre of the earth (in $; N$)?

MediumNEET 2019

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

The gravitational force of the Earth on a rocket at a height $h$ from the surface of the Earth is $\frac{1}{3}$ of the gravitational force on the surface of the Earth. Find the relationship between $h$ and $R_e$ (the radius of the Earth).

Similar Questions

Derive the equation for the variation of $g$ due to height from the surface of the Earth.

If the change in the value of $g$ at a height $h$ above the surface of the earth is the same as at a depth $x$ below it,then (where $x$ and $h$ are much smaller than the radius of the earth $R_e$):

The acceleration due to gravity on the Moon is $\frac{1}{6}$ times that of the Earth. If the ratio of the density of the Earth $(\rho_e)$ to the Moon $(\rho_m)$ is $\frac{\rho_e}{\rho_m} = \frac{5}{3}$,what is the radius of the Moon $(R_m)$ in terms of the radius of the Earth $(R_e)$?

The acceleration due to gravity on the surface of the Earth is $g$. What is the acceleration due to gravity at a height $h = R$ above the Earth's surface,where $R$ is the radius of the Earth?

$A$ body weighs $200 \; N$ on the surface of the earth. How much will it weigh halfway down to the centre of the earth (in $; N$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module