(D) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{4}{3}\pi \rho GR$,where $\rho$ is the density,$G$ is the gra

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

(D) The acceleration due to gravity $g$ at the surface of a planet is given by $g = \frac{4}{3}\pi \rho GR$,where $\rho$ is the density,$G$ is the gravitational constant,and $R$ is the radius of the planet.Given that $g_p = g_e$ and $\rho_p = 2\rho_e$,we have:$\frac{g_p}{g_e} = \frac{\rho_p R_p}{\rho_e R_e} = 1$Substituting the given values:$1 = \frac{2\rho_e R_p}{\rho_e R_e}$$1 = 2 \frac{R_p}{R_e}$$R_p = \frac{R_e}{2} = \frac{R}{2}$Thus,the radius of the planet is $\frac{R}{2}$.

Class 11 Physics Gravitation Acceleration Due to Gravity and its Variation

Medium

The density of a newly discovered planet is twice that of Earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the Earth. If the radius of the Earth is $R$,the radius of the planet would be

AIPMT 2004 All AIPMT

A
$2R$
B
$4R$
C
$\frac{1}{4}R$
D
$\frac{1}{2}R$

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

Previous Year

AIPMT 2004 Paper All AIPMT years →

Two planets $A$ and $B$ of radii $R$ and $1.5 R$ have densities $\rho$ and $\rho / 2$ respectively. The ratio of acceleration due to gravity at the surface of $B$ to $A$ is:

EasyJEE MAIN 2023

View Solution

As we move from the poles of the Earth towards the equator,what happens to the value of the acceleration due to gravity $(g)$?

Easy

View Solution

The time period of a simple pendulum on the surface of the earth is $T$. If the pendulum is taken to a height equal to half of the radius of the earth,then its time period is

MediumAP EAMCET 2025

View Solution

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$,the radius of the earth,is

MediumAIEEE 2009

View Solution

What is the value of acceleration due to gravity $(g)$ at the centre of the Earth? What will be the variation of $g$ below and above the surface of the Earth?

Easy

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 density of a newly discovered planet is twice that of Earth. The acceleration due to gravity at the surface of the planet is equal to that at the surface of the Earth. If the radius of the Earth is $R$,the radius of the planet would be

Similar Questions

Two planets $A$ and $B$ of radii $R$ and $1.5 R$ have densities $\rho$ and $\rho / 2$ respectively. The ratio of acceleration due to gravity at the surface of $B$ to $A$ is:

As we move from the poles of the Earth towards the equator,what happens to the value of the acceleration due to gravity $(g)$?

The time period of a simple pendulum on the surface of the earth is $T$. If the pendulum is taken to a height equal to half of the radius of the earth,then its time period is

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where $g$ = the acceleration due to gravity on the surface of the earth) in terms of $R$,the radius of the earth,is

What is the value of acceleration due to gravity $(g)$ at the centre of the Earth? What will be the variation of $g$ below and above the surface of the Earth?

Vedclass Test Series

Exam Paper Generator

Online Exam Module