Gravitation

JEE Main level, single correct, medium

152 of 223

Question 152 of 223



Two hypothetical planets of masses m1\mathrm{m}_1 and m2\mathrm{m}_2 are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres. What is their speed when their separation is ' dd '?
(Speed of m1\mathrm{m}_1 is v1\mathrm{v}_1 and that of m2\mathrm{m}_2 is v2\mathrm{v}_2 )

Show answer and solution

Answer: (B) v1=m22Gd(m1+m2)v2=m12Gd(m1+m2)\begin{aligned} &\mathrm{v}_1=\mathrm{m}_2 \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)} \\ &\mathrm{v}_2=\mathrm{m}_1 \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)} \end{aligned}

We choose reference point, infinity, where total energy of the system is zero.
So, initial energy of the system =0=0
Final energy
=12 m1v12+12 m2v22−Gm1 m2 d=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1^2+\frac{1}{2} \mathrm{~m}_2 \mathrm{v}_2^2-\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}
From conservation of energy, Initial energy == Final energy
∴0=12 m1v12+12 m2v22−Gm1 m2 d\therefore 0=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1{ }^2+\frac{1}{2} \mathrm{~m}_2 \mathrm{v}_2{ }^2-\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}
or 12 m1v12+12 m1v22=Gm1 m2 d…(1)\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1^2+\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_2^2=\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}} \ldots(1)
By conservation of linear momentum
m1v1+m2v2=0\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2=0
or v1v2=−m2 m1⇒v2=−m1 m2v1\frac{\mathrm{v}_1}{\mathrm{v}_2}=-\frac{\mathrm{m}_2}{\mathrm{~m}_1} \Rightarrow \mathrm{v}_2=-\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{v}_1
Putting value of v2\mathrm{v}_2 in equation (1), we get
m1v12+m2(−m1v1 m2)2=2Gm1 m2 d\mathrm{m}_1 \mathrm{v}_1^2+\mathrm{m}_2\left(-\frac{\mathrm{m}_1 \mathrm{v}_1}{\mathrm{~m}_2}\right)^2=\frac{2 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}

m1m2v12+m12v12m2=2Gm1m2 dv1=2Gm22 d( m1+m2)=m22Gd(m1+m2) Similarly v2=−m12Gd(m1+m2)\begin{aligned} &\frac{m_1 m_2 v_1^2+m_1^2 v_1^2}{m_2}=\frac{2 \mathrm{Gm}_1 m_2}{\mathrm{~d}} \\ &\mathrm{v}_1=\sqrt{\frac{2 \mathrm{Gm}_2^2}{\mathrm{~d}\left(\mathrm{~m}_1+\mathrm{m}_2\right)}}=\mathrm{m}_2 \sqrt{\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right)}} \\ &\text { Similarly } \mathrm{v}_2=-\mathrm{m}_1 \sqrt{\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right)}} \end{aligned}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Gravitation and orbits simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Gravitation (Gravitational potential and Potential Energy), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.