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
- A$2R$
- B$4R$
- C$\frac{1}{4}R$
- D$\frac{1}{2}R$
Explore More
Similar Questions
Obtain an expression for the acceleration due to gravity of the Earth at a depth $d$ below its surface.
Difficult
View SolutionMatch Column-$I$ with Column-$II$.
Column-$I$ Column-$II$ $(1)$ Maximum value of acceleration due to gravity $g$ $(a)$ At the center of the Earth $(2)$ Minimum value of acceleration due to gravity $g$ $(b)$ At the poles $(3)$ Zero value of acceleration due to gravity $g$ $(c)$ At the equator
| Column-$I$ | Column-$II$ |
| $(1)$ Maximum value of acceleration due to gravity $g$ | $(a)$ At the center of the Earth |
| $(2)$ Minimum value of acceleration due to gravity $g$ | $(b)$ At the poles |
| $(3)$ Zero value of acceleration due to gravity $g$ | $(c)$ At the equator |
Medium
View SolutionThe acceleration due to gravity at height $h$ above the earth if $h \ll R$ (radius of earth) is given by
$A$ $90 \,kg$ body placed at $2R$ distance from the surface of the Earth experiences a gravitational pull of: ($R=$ Radius of Earth,$g=10 \,ms^{-2}$) (in $\,N$)
DifficultJEE MAIN 2024
View SolutionThe radius of the Earth is approximately $6000 \, km$. The weight of a body at a height of $6000 \, km$ from the Earth's surface becomes:
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo