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JEE Main 20238 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2023 · 8 April · Evening Shift · Physics · Gravitation

The acceleration due to gravity at height h above the earth if h ≪...

The acceleration due to gravity at height hh above the earth if h≪Rh ≪ R (Radius of earth) is given by

Options

  1. Ag′=g(1−h22R2){g}^{'} = g\left(1 - \dfrac{{h}^{2}}{2{R}^{2}}\right)
  2. Bg′=g(1−h2R){g}^{'} = g\left(1 - \dfrac{h}{2R}\right)
  3. Cg′=g(1−2hR){g}^{'} = g\left(1 - \dfrac{2h}{R}\right)
  4. Dg′=g(1−2h2R2){g}^{'} = g\left(1 - \dfrac{2{h}^{2}}{{R}^{2}}\right)

Correct answer

C. g′=g(1−2hR){g}^{'} = g\left(1 - \dfrac{2h}{R}\right)

Step-by-step solution

The acceleration due to gravity g′g ' at a height hh from the surface of Earth is given by

g′=GM(R+h)2...(1){g}^{'} = \dfrac{GM}{( R + h{)}^{2}} . . . \left(1\right)

Use the method of binomial expansion to obtain the approximate expression for the acceleration due to gravity at a particular height from the surface of the Earth.

{g′=GMR2(1+hR)2=GMR2(1+hR)−2≈GMR2(1−2hR)...(2)\begin{cases}{g}^{'} & = & \dfrac{GM}{{R}^{2}{\left(1 + \dfrac{h}{R}\right)}^{2}} \\ & = & \dfrac{GM}{{R}^{2}}{\left(1 + \dfrac{h}{R}\right)}^{- 2} \\ & \approx & \dfrac{GM}{{R}^{2}}\left(1 - \dfrac{2h}{R}\right) . . . \left(2\right)\end{cases}

The formula to calculate the acceleration due to gravity at the surface of the Earth is given by

g=GMR2...(3)g = \dfrac{GM}{{R}^{2}} . . . ( 3 )

Divide equation (2) by equation (3) and simplify to obtain the required relation.

{g′g=GMR2(1−2hR)GMR2=(1−2hR)⇒g′=g(1−2hR)\begin{cases}\dfrac{g '}{g} & = & \dfrac{\dfrac{GM}{{R}^{2}}\left(1 - \dfrac{2h}{R}\right)}{\dfrac{GM}{{R}^{2}}} \\ & = & \left(1 - \dfrac{2h}{R}\right) \\ & ⇒ & g ' = g\left(1 - \dfrac{2h}{R}\right)\end{cases}

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