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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Physics
Chapter
Gravitation

JEE Main 2024 · 4 April · Evening Shift · Physics · Gravitation

Correct formula for height of a satellite from earths surface is:

Correct formula for height of a satellite from earths surface is:Correct formula for height of a satellite from earths surface is:

Options

  1. A(T2R24π2g)1/3−R\left(\frac{T^2 R^2}{4 \pi^2 g}\right)^{1 / 3}-R
  2. B(T2R2g4π2)1/3−R\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{1 / 3}-R
  3. C(T2R2g4π2)−1/3+R\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{-1 / 3}+R
  4. D(T2R2g4π)1/2−R\left(\frac{T^2 R^2 g}{4 \pi}\right)^{1 / 2}-R

Correct answer

B. (T2R2g4π2)1/3−R\left(\frac{T^2 R^2 g}{4 \pi^2}\right)^{1 / 3}-R

Step-by-step solution

Correct formula for height of a satellite from earths surface is:
⇒GMm(R+h)2=mv2(R+h)⇒GM(R+h)=v2\Rightarrow \frac{\mathrm{GMm}}{(\mathrm{R}+\mathrm{h})^2}=\frac{\mathrm{mv}^2}{(\mathrm{R}+\mathrm{h})} \Rightarrow \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})}=\mathrm{v}^2 ....(1) ⇒v=(R+h)ω⇒v=(R+h)2πT\Rightarrow \mathrm{v}=(\mathrm{R}+\mathrm{h}) \omega \Rightarrow \mathrm{v}=(\mathrm{R}+\mathrm{h}) \frac{2 \pi}{\mathrm{T}} ....(2) ⇒GMR2=g⇒\Rightarrow \frac{\mathrm{GM}}{\mathrm{R}^2}=\mathrm{g} \Rightarrow\mathrm{GM}=\mathrm{gR}2^2 ....(3) Put value from (2) \& (3) in eq. (1) ⇒gR2(R+h)=(R+h)2(2πT)2⇒T2R2 g(2π)2=(R+h)3⇒[T2R2 g(2π)2]1/3−R=h\begin{aligned} & \Rightarrow \frac{\mathrm{gR}^2}{(\mathrm{R}+\mathrm{h})}=(\mathrm{R}+\mathrm{h})^2\left(\frac{2 \pi}{\mathrm{T}}\right)^2 \\ & \Rightarrow \frac{\mathrm{T}^2 \mathrm{R}^2 \mathrm{~g}}{(2 \pi)^2}=(\mathrm{R}+\mathrm{h})^3 \\ & \Rightarrow\left[\frac{\mathrm{T}^2 \mathrm{R}^2 \mathrm{~g}}{(2 \pi)^2}\right]^{1 / 3}-\mathrm{R}=\mathrm{h} \end{aligned}
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