JEE Challenger
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PhyEasyAdv 2023 • Paper 1

Ratio of Gravitational Accelerations of Two Satellites in Circular Orbits

Two satellites P\text{P} and Q\text{Q} are moving in different circular orbits around the Earth (radius RR). The heights of P\text{P} and Q\text{Q} from the Earth surface are hPh_{\text{P}} and hQh_{\text{Q}}, respectively, where hP=R/3h_{\text{P}} = R/3. The accelerations of P\text{P} and Q\text{Q} due to Earth's gravity are gPg_{\text{P}} and gQg_{\text{Q}}, respectively. If gP/gQ=36/25g_{\text{P}}/g_{\text{Q}} = 36/25, what is the value of hQh_{\text{Q}}?

Options

A

3R/53R/5

Correct
B

R/6R/6

C

6R/56R/5

D

5R/65R/6

Topics & Concepts

GravitationGravitation

Step-by-Step Solution

The acceleration due to Earth's gravity at a distance r=R+hr = R + h from the center of the Earth is given by: g(h)=GM(R+h)2g(h) = \frac{GM}{(R + h)^2}

where:

  • GG is the universal gravitational constant,
  • MM is the mass of the Earth,
  • RR is the radius of the Earth, and
  • hh is the height of the satellite above the Earth's surface.

For the two satellites P\text{P} and Q\text{Q} at heights hPh_{\text{P}} and hQh_{\text{Q}} respectively: gP=GM(R+hP)2g_{\text{P}} = \frac{GM}{(R + h_{\text{P}})^2} gQ=GM(R+hQ)2g_{\text{Q}} = \frac{GM}{(R + h_{\text{Q}})^2}

Taking the ratio of their gravitational accelerations: gPgQ=(R+hQ)2(R+hP)2\frac{g_{\text{P}}}{g_{\text{Q}}} = \frac{(R + h_{\text{Q}})^2}{(R + h_{\text{P}})^2}

Given that gPgQ=3625\frac{g_{\text{P}}}{g_{\text{Q}}} = \frac{36}{25} and hP=R3h_{\text{P}} = \frac{R}{3}: (R+hQ)2(R+R3)2=3625\frac{(R + h_{\text{Q}})^2}{\left(R + \frac{R}{3}\right)^2} = \frac{36}{25}

Taking the positive square root on both sides: R+hQ4R3=65\frac{R + h_{\text{Q}}}{\frac{4R}{3}} = \frac{6}{5}

Multiplying both sides by 4R3\frac{4R}{3}: R+hQ=65×4R3=8R5R + h_{\text{Q}} = \frac{6}{5} \times \frac{4R}{3} = \frac{8R}{5}

Solving for hQh_{\text{Q}}: hQ=8R5−R=3R5h_{\text{Q}} = \frac{8R}{5} - R = \frac{3R}{5}

Therefore, the correct option is (A).

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