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Acceleration due to gravity below and above the surface of earth

The height at which the acceleration due to gravity becomes {g

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Solving time: 3 mins

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where, $g$ is the acceleration due to gravity on the surface of the earth) in terms of $R$ , the radius of the earth is
(a) $2 R$
(b) $\frac{R}{3}$
(c) $\frac{R}{2}$
(d) $2 R$

Name: Video solution 1: The height at which the acceleration due to gravity becomes 9g (where, g is the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is(a) 2R(b) 3R(c) 2R(d) 2R
Uploaded: 2023-03-28T08:58:42.000Z
Duration: 3 min
Description: Video solution 1: The height at which the acceleration due to gravity becomes 9g (where, g is the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is(a) 2R(b) 3R(c) 2R(d) 2R

Views: 6,449 students

Updated on: Mar 28, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2009

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Sofia Discussed

The height at which the acceleration due to gravity becomes

\frac{g}{9}

(where,

g

is the acceleration due to gravity on the surface of the earth) in terms of

R

, the radius of the earth is
(a)

2 R

(b)

\frac{R}{3}

(c)

\frac{R}{2}

(d)

2 R

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Exp. (a)
As value of acceleration due to gravity above the earth surface is

g

g^{'} = \frac{GM}{( R + h ) ^{2}}

, acceleration due to gravity at height

h

\Rightarrow \frac{g}{9} = \frac{GM}{R ^{2}} \cdot \frac{R ^{2}}{( R + h ) ^{2}} = g (\frac{R}{R + h})^{2}

\Rightarrow \frac{1}{9} = (\frac{R}{R + h})^{2} \Rightarrow \frac{R}{R + h} = \frac{1}{3}

\Rightarrow 3 R = R + h \Rightarrow 2 R = h

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Question Text	The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where, $g$ is the acceleration due to gravity on the surface of the earth) in terms of $R$ , the radius of the earth is (a) $2 R$ (b) $\frac{R}{3}$ (c) $\frac{R}{2}$ (d) $2 R$
Updated On	Mar 28, 2023
Topic	Gravitation
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 2
Upvotes	236
Avg. Video Duration	4 min

The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where, $g$ is the acceleration due to gravity on the surface of the earth) in terms of $R$ , the radius of the earth is
(a) $2 R$
(b) $\frac{R}{3}$
(c) $\frac{R}{2}$
(d) $2 R$

Text solutionVerified

Filo tutor solutions (2)

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The height at which the acceleration due to gravity becomes 9g​ (where, g is the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth is(a) 2R(b) 3​R​(c) 2R​(d) 2​R

Text solutionVerified

Filo tutor solutions (2)

Questions from JEE Mains 2009 - PYQs

Practice questions from Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

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The height at which the acceleration due to gravity becomes $\frac{g}{9}$ (where, $g$ is the acceleration due to gravity on the surface of the earth) in terms of $R$ , the radius of the earth is
(a) $2 R$
(b) $\frac{R}{3}$
(c) $\frac{R}{2}$
(d) $2 R$