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Physics · Gravitation

JEE Main 2024 — 5 April, Shift 2 — Question 48

A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is : (( Given == Radius of geo-stationary orbit for earth is 4.2×104 km4.2 \times 10^{4} \mathrm{~km} )

  1. Option A:

    1.4×104 km1.4 \times 10^{4} \mathrm{~km}

  2. Option B:

    8.4×104 km8.4 \times 10^{4} \mathrm{~km}

  3. Option C:

    1.68×105 km1.68 \times 10^{5} \mathrm{~km}

  4. Option D:

    1.05×104 km1.05 \times 10^{4} \mathrm{~km}

    Correct

Answer: D

Step-by-step solution

T=2πr3/2GM\mathrm{T}=\frac{2 \pi \mathrm{r}^{3 / 2}}{\sqrt{\mathrm{GM}}}

T1 T2=(r1r2)3/2(M2M1)1/2\frac{\mathrm{T}_{1}}{\mathrm{~T}_{2}}=\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)^{3 / 2}\left(\frac{\mathrm{M}_{2}}{\mathrm{M}_{1}}\right)^{1 / 2}

624=(r1)3/2(4.2×104)3/2(MM/4)1/2\frac{6}{24}=\frac{\left(r_{1}\right)^{3 / 2}}{\left(4.2 \times 10^{4}\right)^{3 / 2}}\left(\frac{M}{M / 4}\right)^{1 / 2}

r1=1.05×104 km\mathrm{r}_{1}=1.05 \times 10^{4} \mathrm{~km}

Answer key and solution verified before publishing.

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