A particle of mass ‘m’ is projected with a velocity v=kVe(k < 1) from the surface of the earth. (Ve=escape velocity) The maximum height above the surface reached by the particle is :

  1. \(\frac{{R{k^2}}}{{1 - {k^2}}}\)
  2. \(R{\left( {\frac{k}{{1 - k}}} \right)^2}\)
  3. \(R{\left( {\frac{k}{{1 + k}}} \right)^2}\)
  4. \({\frac{{{R^2}k}}{{1 + k}}}\)

Answer (Detailed Solution Below)

Option 1 : \(\frac{{R{k^2}}}{{1 - {k^2}}}\)
Free
CT 1: Botany (Cell:The Unit of Life)
30.5 K Users
25 Questions 100 Marks 20 Mins

Detailed Solution

Download Solution PDF

Concept:

The minimum velocity required by a body to overcome the gravitational force of earth and escape from earth so that it never falls back is called as escape velocity.

Escape velocity formula is given

\({V} = \sqrt {\frac{{2GM}}{R}} \)

Where, V is the escape velocity

G is the gravitational constant is 6.67 × 10-11 m3 kg-1 s-2

M is the mass of the planet

R is the radius from the center of gravity

Calculation:

Given v = kVe

where, k < 1

Thus, v < Ve

From conservation of mechanical energy,

\(\frac{1}{2}m{v^2} - \frac{{GmM}}{R} = - \frac{{GmM}}{{(R + h)}}\)

\( \Rightarrow \frac{{{V^2}}}{2} = \frac{{GM}}{R} - \frac{{(GM)}}{{(R + h)}} = \frac{h}{{R(R + h)}}GM\)

\( \Rightarrow \frac{1}{2}{k^2}V_e^2 = \frac{{GMh}}{{R(R + h)}}\)

We know, \({V_e} = \sqrt {\frac{{2GM}}{R}} \)

\( \Rightarrow \frac{1}{2}{k^2}\left( {\frac{{2GM}}{R}} \right) = \frac{{GMh}}{{R(R + h)}}\)

\({k^2} = \frac{h}{{(R + h)}}\)

Rk2 + hk2 = h

Rk2 = h(1 – k2)

\(\therefore h = \frac{{R{k^2}}}{{(1 - {k^2})}}\)

Latest NEET Updates

Last updated on Jun 3, 2025

->NEET provisional answer key 2025 was made available on June 3, 2025 on the official website for the students to check.

->NEET 2025 exam is over on May 4, 2025.

-> The NEET 2025 Question Papers PDF are now available.

-> NTA has changed the NEET UG Exam Pattern of the NEET UG 2025. Now, there will be no Section B in the examination.

-> Candidates preparing for the NEET Exam, can opt for the latest NEET Mock Test 2025

-> NEET aspirants can check the NEET Previous Year Papers for their efficient preparation. and Check NEET Cut Off here.

More Escape speed Questions

More Gravitation Questions

Get Free Access Now
Hot Links: teen patti royal teen patti game paisa wala teen patti real