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Gravitation
Escape velocity
4 questions from 4 papers, each with its answer.
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- 4
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- years asked, 2017–2026
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- of Gravitation
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Appears in
Concepts tested
- Escape velocity
- Escape velocity from a height
- escape-velocity-density-radius
- Schwarzschild radius
All Questions
2024 Re-NEET 1 question
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The escape velocity for earth is $v$. A planet having 9 times mass that of earth and radius, 16 times that of earth, has the escape velocity of:$V_e=\sqrt{\dfrac{2GM}{R}}\propto\sqrt{\dfrac{M}{R}}$. $\dfrac{(V_e)_p}{(V_e)_e}=\sqrt{\dfrac{M_p}{M_e}\times\dfrac{R_e}{R_p}}=\sqrt{\dfrac{9}{16}}=\dfrac{3}{4}$, so $(V_e)_p=\dfrac{3}{4}v$.
2016 1 question
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The ratio of escape velocity at earth ($v_e$) to the escape velocity at a planet ($v_p$) whose radius and mean density are twice as that of earth is
$v_e \propto R\sqrt{\rho}$, so doubling both gives $2\sqrt{2}$ times.
Escape velocity $V = \sqrt{\frac{2GM}{R}}$, with $M = \rho\times\frac{4}{3}\pi R^3$
For the planet: $R_p = 2R_e$, $\rho_p = 2\rho_e$, so $M_p = 2\rho_e\times\frac{4}{3}\pi(2R_e)^3 = 16M_e$
$V_p = \sqrt{\frac{2G(16M_e)}{2R_e}} = \sqrt{\frac{2GM_e}{R_e}\times8} = 2\sqrt{2}V_e$
$V_e : V_p = 1 : 2\sqrt{2}$
2014 AIPMT 1 question
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A black hole is an object whose gravitational field is so strong that even light cannot escape from it. To what approximate radius would earth (mass = $5.98\times10^{24}$ kg) have to be compressed to be a black hole?Escape velocity $= \sqrt{\frac{2GM}{R}} = c$ (speed of light)
$R = \frac{2GM}{c^2} = \frac{2\times 6.6\times10^{-11}\times 5.98\times10^{24}}{(3\times10^8)^2}$ m
$R \approx 10^{-2}$ m
2006 AIPMT-PRE 1 question
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The Earth is assumed to be a sphere of radius $R$. A platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$, where $v$ is its escape velocity from the surface of the Earth. The value of $f$ is :-The platform is at a distance $R + R = 2R$ from the centre of the Earth.
By conservation of mechanical energy, for the body to just escape (zero energy at infinity): $-\dfrac{GMm}{R + R} + \dfrac{1}{2}m(fv)^2 = 0 + 0$
$\dfrac{1}{2}(fv)^2 = \dfrac{GM}{2R}$
$\Rightarrow fv = \sqrt{\dfrac{GM}{R}}$
But the escape velocity from the surface is $v = \sqrt{\dfrac{2GM}{R}}$
Therefore $f\sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{GM}{R}}$
$\Rightarrow f = \dfrac{1}{\sqrt{2}}$