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JEE Advanced · Physics · 9. Gravitation

Two spherical planets \(P\) and \(Q\) have the same uniform density \(\rho\), masses \(M_{P}\) and \(M_{Q}\) and surface areas \(A\) and 4A respectively. A spherical planet \(R\) also has uniform density \(\rho\) and its mass is \(\left(M_{P}+M_{Q}\right)\). The escape velocities from the planets \(P, Q\) and \(R\) are \(V_{P}, V_{Q}\) and \(V_{R}\), respectively. Then

  1. A \(V_{Q}>V_{R}>V_{P}\)
  2. B \(V_{R}>V_{Q}>V_{P}\)
  3. C \(V_{R} / V_{P}=3\)
  4. D \(V_{P} / V_{Q}=\frac{1}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(V_{P} / V_{Q}=\frac{1}{2}\)

Step-by-step Solution

Detailed explanation

Here Planets \(P\) and \(Q\) have the same uniform density ' \(\rho\) ' and surface areas \(A\) and \(4 A\) respectively. Let the mass of \(P, M_{P}\) be \(m\).

Then \(m=\rho \times \frac{4}{3} \pi r^{3}=\rho \times \frac{4}{3} \pi\left[\frac{A}{4 \pi}\right]^{3 / 2}\)

The mass of \(M_{Q}=\rho \times \frac{4}{3} \pi\left[\frac{4 A}{4 \pi}\right]^{3 / 2}=8 \mathrm{~m}\)

\(\therefore \quad\) The mass of Planet \(R=8 \mathrm{~m}+\mathrm{m}=9 \mathrm{~m}\)

If the radius of \(P=r\)

Then the radius of \(Q=2 r\)

\(\left[\because r_{Q}=\left(\frac{4 A}{4 \pi}\right)^{3 / 2}=2\left(\frac{A}{4 \pi}\right)^{3 / 2}\right]\)

and radius of \(R=9^{1 / 3} r\)

\(\left[\begin{array}{l}

\because M_{R}=M_{P}+M_{Q} \\

r_{R}^{3}=r^{3}+(2 r)^{3}=9 r^{3}

\end{array}\right]\)

As we know, escape velocity from the planet

\(V_{e}=\sqrt{\frac{2 G M}{R}} \therefore v_{P}=\sqrt{\frac{2 G M_{P}}{R_{p}}}=\sqrt{\frac{2 G m}{r}} \)
\( v_{Q} =\sqrt{\frac{2 G M_{Q}}{R_{Q}}}=\sqrt{\frac{2 G(8 \mathrm{~m})}{2 r}}=2 v_{P} \)
\( v_{R} =\sqrt{\frac{2 G(9 \mathrm{~m})}{9^{1 / 3} r}}=9^{1 / 3} v_{P}\)
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