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Physics · Gravitation

JEE Main 2024 — 4 April, Shift 2 — Question 37

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

  1. Option A:

    (T2R2g4π)1/2−R\left(\frac{T^{2} R^{2} g}{4 \pi}\right)^{1 / 2}-R

  2. Option B:

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

    Correct
  3. Option C:

    (T2R24π2g)1/3−R\left(\frac{T^{2} R^{2}}{4 \pi^{2} g}\right)^{1 / 3}-R

  4. Option D:

    (T2R24π2)−1/3+R\left(\frac{T^{2} R^{2}}{4 \pi^{2}}\right)^{-1 / 3}+R

Answer: B

Step-by-step solution

⇒GMm(R+h)2=mv2(R+h)\Rightarrow \frac{\text{GMm}}{{{(\text{R}+\text{h})}^{2}}}=\frac{\text{m}{{\text{v}}^{2}}}{\left( \text{R}+\text{h} \right)} ⇒GM(R+h)=v2\Rightarrow \frac{GM}{\left( R+h \right)}={{v}^{2}}. ⇒v=(R+h)ω\Rightarrow \text{v}=\left( \text{R}+\text{h} \right)\omega ⇒v=(R+h)2π  ⁣ ⁣  ⁣ ⁣ T\Rightarrow \text{v}=\left( \text{R}+\text{h} \right)\frac{2\pi }{\text{ }\!\!~\!\!\text{ T}}. ⇒GMR2=g\Rightarrow \frac{\text{GM}}{{{\text{R}}^{2}}}=\text{g} ⇒GM=gR2\Rightarrow \text{GM}=\text{g}{{\text{R}}^{2}} Put value from (2) & (3) in eq. (1) ⇒gR2(R+h)=(R+h)2(2π  ⁣ ⁣  ⁣ ⁣ T)2\Rightarrow \frac{\text{g}{{\text{R}}^{2}}}{\left( \text{R}+\text{h} \right)}={{(\text{R}+\text{h})}^{2}}{{\left( \frac{2\pi }{\text{ }\!\!~\!\!\text{ T}} \right)}^{2}} ⇒T2R2  ⁣ ⁣  ⁣ ⁣ g(2π)2=(R+h)3\Rightarrow \frac{{{\text{T}}^{2}}{{\text{R}}^{2}}\text{ }\!\!~\!\!\text{ g}}{{{(2\pi )}^{2}}}={{(\text{R}+\text{h})}^{3}} ⇒[T2R2  ⁣ ⁣  ⁣ ⁣ g(2π)2]1/3−R=h\Rightarrow {{\left[ \frac{{{\text{T}}^{2}}{{\text{R}}^{2}}\text{ }\!\!~\!\!\text{ g}}{{{(2\pi )}^{2}}} \right]}^{1/3}}-\text{R}=\text{h}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
4 April, Shift 2
Subject
Physics
Chapter
Gravitation
Topic
Motion of Satellites and Escape Speed

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