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Gravitation question

2021 · 31 Aug · Shift 1 · Q52
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  5. /2021 · 31 Aug · Shift 1 · Q52

Gravitation question

2021 · 31 Aug · Shift 1 · Q52

JEE MainPhysicsGravitationMCQ+4 / −1
The masses and radii of the earth and moon are (M1, R1) and (M2, R2) respectively. Their centres are at a distance 'r' apart. Find the minimum escape velocity for a particle of mass 'm' to be projected from the middle of these two masses :
  1. A
    V=124G(M1+M2)rV = {1 \over 2}\sqrt {{{4G({M_1} + {M_2})} \over r}}V=21​r4G(M1​+M2​)​​
  2. B
    V=4G(M1+M2)rV = \sqrt {{{4G({M_1} + {M_2})} \over r}}V=r4G(M1​+M2​)​​
  3. C
    V=122G(M1+M2)rV = {1 \over 2}\sqrt {{{2G({M_1} + {M_2})} \over r}}V=21​r2G(M1​+M2​)​​
  4. D
    V=2G(M1+M2)rV = {{\sqrt {2G} ({M_1} + {M_2})} \over r}V=r2G​(M1​+M2​)​
View written solutionFree

Correct answer: B

  1. Point of projection

The particle is projected from the midpoint of the line joining the centres of Earth and Moon.

So its distance from each centre is r2.\frac{r}{2}.2r​.

  1. Gravitational potential at the midpoint

The gravitational potential due to a mass MMM at distance ddd is Vg=−GMd.V_g=-\frac{GM}{d}.Vg​=−dGM​.

Hence, total gravitational potential at the midpoint is V=−GM1r/2−GM2r/2.V= -\frac{GM_1}{r/2}-\frac{GM_2}{r/2}.V=−r/2GM1​​−r/2GM2​​.

Simplifying,

= -\frac{2G(M_1+M_2)}{r}.$$ --- 3. **Potential energy of the particle** For a particle of mass $m$, $$U=mV=-\frac{2Gm(M_1+M_2)}{r}.$$ --- 4. **Condition for minimum escape velocity** For minimum escape velocity, total mechanical energy must be zero at launch: $$\frac{1}{2}mv^2+U=0.$$ So, $$\frac{1}{2}mv^2-\frac{2Gm(M_1+M_2)}{r}=0.$$ Cancelling $m$, $$\frac{1}{2}v^2=\frac{2G(M_1+M_2)}{r}.$$ Therefore, $$v^2=\frac{4G(M_1+M_2)}{r}.$$ Hence, $$v=\sqrt{\frac{4G(M_1+M_2)}{r}}.$$ --- 5. **Match with options** This matches: $$\boxed{\text{B}}$$ since option B is $$V=\sqrt{\frac{4G(M_1+M_2)}{r}}.$$
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