Verified Solution • 2025–26 Session moderate

QJEEAdv23Ph16a5: JEE Advanced 2023 Physics - Class 11 JEEAdv23Ph16a5 Solution: Two satellites and are moving in different circular orbits…

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

Multiple Choice Options

  • A 3R5\frac{3 R}{5}
  • B R6\frac{R}{6}
  • C 6R5\frac{6 R}{5}
  • D 5R5\frac{5 R}{5}

Step-by-Step Verified Solution

The formula to calculate the acceleration due to gravity at a height ( h ) from the surface of the Earth is expressed as :

g=GMe(Re+h)2g = \frac{G M_e}{(R_e + h)^2}

To find the ratio of gravitational acceleration at two different heights ( h_P ) and ( h_Q ) above the Earth's surface, use the formula for each and form a ratio :

gPgQ=GMe(Re+hP)2GMe(Re+hQ)2=(Re+hQ)2(Re+hP)2\frac{g_P}{g_Q} = \frac{\frac{G M_e}{(R_e + h_P)^2}}{\frac{G M_e}{(R_e + h_Q)^2}} = \frac{(R_e + h_Q)^2}{(R_e + h_P)^2}

Given hP=Re3h_P = \frac{R_e}{3} , the ratio simplifies to:

gPgQ=(Re+hQ)2(Re+Re3)2=3625\frac{g_P}{g_Q} = \frac{(R_e + h_Q)^2}{(R_e + \frac{R_e}{3})^2} = \frac{36}{25}

Solving for hQh_Q in terms of ReR_e yields hQ=3Re5h_Q = \frac{3R_e}{5} .

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