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Acceleration due to gravity below and above the surface of earth

A planet of radius R=1 / 10 ( radius of earth) has the same ma

Question

Hard

Solving time: 4 mins

A planet of radius $R = 1/10 \times ($ radius of earth) has the same mass density as earth. Scientists dig a well of depth $\frac{R}{5}$ on it and lower a wire of the same length and of linear mass density $1 0^{- 3} kgm^{- 1}$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of earth $= 6 \times 1 0^{6} m$ and the acceleration due to gravity of earth is $10 ms^{- 2}$ )

96 N

108 N

120 N

150 N

Views: 6,094 students

Updated on: May 14, 2023

Book:

Mechanics Volume 2 (DC Pandey)

Found 7 tutors discussing this question

Olivia Discussed

A planet of radius

R = 1/10 \times (

radius of earth) has the same mass density as earth. Scientists dig a well of depth

\frac{R}{5}

on it and lower a wire of the same length and of linear mass density

1 0^{- 3} kgm^{- 1}

into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of earth

= 6 \times 1 0^{6} m

and the acceleration due to gravity of earth is

10 ms^{- 2}

)

5 mins ago

Discuss this question LIVE

5 mins ago

Text solutionVerified

(b) Given,

R_{planet} = \frac{R _{earth}}{10}

and density,

ρ = \frac{M _{earth}^{10}}{\frac{4}{3} π R _{earth}^{3}} = \frac{M _{planet}}{\frac{4}{3} π R _{planet}^{3}}

\Rightarrow M_{planet} = \frac{M _{earth}}{1 0 ^{3}}

g_{surface of planet} = \frac{G M _{planet}}{R _{planet}^{2}} = \frac{G M _{e} \cdot 1 0 ^{2}}{1 0 ^{3} \cdot R _{e}^{2}} = \frac{G M _{e}}{10 R _{e}^{2}} = \frac{g _{surface of earth}}{10}

g_{depth of planet} = g_{surface of planet} (\frac{x}{R})

where,

x =

distance from centre of planet.

∴

Total force on wire

F = \int_{4 R /5}^{R} λ d xg (\frac{x}{R}) = \frac{λ g}{R} [\frac{x ^{2}}{2}]_{4 R /5}^{R}

Here,

g = g_{surface of planet}

R = R_{planet}

Substituting the given values, we get

F = 108 N

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Question Text	A planet of radius $R = 1/10 \times ($ radius of earth) has the same mass density as earth. Scientists dig a well of depth $\frac{R}{5}$ on it and lower a wire of the same length and of linear mass density $1 0^{- 3} kgm^{- 1}$ into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of earth $= 6 \times 1 0^{6} m$ and the acceleration due to gravity of earth is $10 ms^{- 2}$ )
Updated On	May 14, 2023
Topic	Gravitation
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 3
Upvotes	313
Avg. Video Duration	10 min

Text solutionVerified

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