(D) The orbital radius of the ball is $r = R + h$,where $R$ is the radius of the Earth $(6400 \, km)$ and $h$ is the height $(9 \, km)$.Since $h \ll R

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

(D) The orbital radius of the ball is $r = R + h$,where $R$ is the radius of the Earth $(6400 \, km)$ and $h$ is the height $(9 \, km)$.Since $h \ll R$,we have $r \approx R$.The acceleration of an object in a circular orbit is given by the centripetal acceleration $a = v^2 / r$.Substituting the orbital velocity $v = \sqrt{GM/r}$,we get $a = (GM/r) / r = GM/r^2$.Since $r \approx R$,the acceleration $a \approx GM/R^2$.By definition,the acceleration due to gravity near the Earth's surface is $g = GM/R^2$.Therefore,the acceleration of the ball in orbit is nearly equal to $g$.

Class 11 Physics Gravitation Acceleration Due to Gravity and its Variation

Medium

$A$ ball is launched from the top of Mt. Everest,which is at an elevation of $9000 \, m$. The ball moves in a circular orbit around the Earth. Acceleration due to gravity near the Earth's surface is $g$. The magnitude of the ball's acceleration while in orbit is

KVPY 2015 All KVPY

A
close to $g/2$
B
zero
C
much greater than $g$
D
nearly equal to $g$

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

KVPY 2015 Paper All KVPY years →

The diameters of two planets are in the ratio $4:1$. Their mean densities have the ratio $1:2$. The ratio of gravitational acceleration on the surface of the planets will be:

Medium

View Solution

If $g_E$ and $g_M$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces,one will find the ratio (electronic charge on the moon / electronic charge on the earth) to be

MediumAIEEE 2007

View Solution

The acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth reduces to half of its original value and mass remains constant,then the acceleration due to gravity on the surface of Earth would be:

DifficultJEE MAIN 2024

View Solution

Given below are two statements:
Statement-$I$: Acceleration due to gravity is different at different places on the surface of Earth.
Statement-$II$: Acceleration due to gravity increases as we go down below the Earth's surface.
In the light of the above statements,choose the correct answer from the options given below:

MediumJEE MAIN 2023

View Solution

Acceleration due to gravity at the surface of a planet is equal to that at the surface of the Earth,and its density is $1.5$ times that of the Earth. If the radius of the Earth is $R$,the radius of the planet is:

Medium

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

$A$ ball is launched from the top of Mt. Everest,which is at an elevation of $9000 \, m$. The ball moves in a circular orbit around the Earth. Acceleration due to gravity near the Earth's surface is $g$. The magnitude of the ball's acceleration while in orbit is

Similar Questions

The diameters of two planets are in the ratio $4:1$. Their mean densities have the ratio $1:2$. The ratio of gravitational acceleration on the surface of the planets will be:

If $g_E$ and $g_M$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces,one will find the ratio (electronic charge on the moon / electronic charge on the earth) to be

The acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth reduces to half of its original value and mass remains constant,then the acceleration due to gravity on the surface of Earth would be:

Acceleration due to gravity at the surface of a planet is equal to that at the surface of the Earth,and its density is $1.5$ times that of the Earth. If the radius of the Earth is $R$,the radius of the planet is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module