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Gravitation question

2023 · Shift 1 · Q41
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  5. /2023 · Shift 1 · Q41

Gravitation question

2023 · Shift 1 · Q41

JEE AdvancedPhysicsGravitationMCQ+3 / −1
Two satellites P\mathrm{P}P and Q\mathrm{Q}Q are moving in different circular orbits around the Earth (radius RRR). The heights of P\mathrm{P}P and Q\mathrm{Q}Q from the Earth surface are hPh_{\mathrm{P}}hP​ and hQh_{\mathrm{Q}}hQ​, respectively, where hP=R/3h_{\mathrm{P}}=R / 3hP​=R/3. The accelerations of P\mathrm{P}P and Q\mathrm{Q}Q due to Earth's gravity are gPg_{\mathrm{P}}gP​ and gQg_{\mathrm{Q}}gQ​, respectively. If gP/gQ=36/25g_{\mathrm{P}} / g_{\mathrm{Q}}=36 / 25gP​/gQ​=36/25, what is the value of hQh_{\mathrm{Q}}hQ​ ?
  1. A
    3R5\frac{3 R}{5}53R​
  2. B
    R6\frac{R}{6}6R​
  3. C
    6R5\frac{6 R}{5}56R​
  4. D
    5R5\frac{5 R}{5}55R​
View written solutionFree

Correct answer: A

  1. Use the formula for gravitational acceleration at height hhh

    At a distance r=R+hr = R+hr=R+h from the Earth's center, g(h)=GM(R+h)2g(h)=\frac{GM}{(R+h)^2}g(h)=(R+h)2GM​

    So for satellites PPP and QQQ, gP=GM(R+hP)2,gQ=GM(R+hQ)2g_P=\frac{GM}{(R+h_P)^2}, \qquad g_Q=\frac{GM}{(R+h_Q)^2}gP​=(R+hP​)2GM​,gQ​=(R+hQ​)2GM​

  2. Form the ratio

    =\frac{(R+h_Q)^2}{(R+h_P)^2}$$ Given, $$\frac{g_P}{g_Q}=\frac{36}{25}$$ and $$h_P=\frac{R}{3}$$

  3. Substitute hPh_PhP​

    R+hP=R+R3=4R3R+h_P = R+\frac{R}{3}=\frac{4R}{3}R+hP​=R+3R​=34R​

    Hence, (R+hQ)2(4R3)2=3625\frac{(R+h_Q)^2}{\left(\frac{4R}{3}\right)^2} = \frac{36}{25}(34R​)2(R+hQ​)2​=2536​

  4. Solve for R+hQR+h_QR+hQ​

    Taking square root on both sides, R+hQ4R/3=65\frac{R+h_Q}{4R/3} = \frac{6}{5}4R/3R+hQ​​=56​

    Therefore, R+hQ=65⋅4R3=24R15=8R5R+h_Q = \frac{6}{5}\cdot \frac{4R}{3} = \frac{24R}{15} = \frac{8R}{5}R+hQ​=56​⋅34R​=1524R​=58R​

  5. Find hQh_QhQ​

    hQ=8R5−R=8R5−5R5=3R5h_Q = \frac{8R}{5}-R = \frac{8R}{5}-\frac{5R}{5} = \frac{3R}{5}hQ​=58R​−R=58R​−55R​=53R​

  6. Check options

    hQ=3R5h_Q=\frac{3R}{5}hQ​=53R​ which matches Option A.

Final Answer: 3R5\boxed{\frac{3R}{5}}53R​​

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