Question

The velocity with which a body should be projected from the surface of the earth such that it reaches a maximum height equal to 'n' times the radius of the earth is
(1) √ n GM /(n+1)R
(2) √n GM /R
(3) √ 2nGM / (n+1)R
(4) √GM / nR ​

Answer 1

Explanation:

The velocity with which a body should be projected from the surface of the earth such that it reaches a maximum height equal to 5 times radius R fo the earth is. (M is mass of the earth ) u=√53GMR.

Answer 2

Given:-

a body is projected from the surface of the earth such that it reaches a maximum height equal to 'n' times the radius of the earth.

To Find:-

• Velocity.

Solution:-

Let v be the Velocity of A Body that is Projected

then we can say that it's Kinetic energy will be= $\frac{1}{2} \times m \times {v}^{2}$

IT is Projected from the surface of the Earth

So,

Kinetic Energy= Increase In Potential Energy.

Now, Potential Energy will be =$\frac{-GMm}{ R}$

We Know that

$GM = g { R}^{2}$

So,

$= > p.e = \frac{ - mg { R}^{2} }{ R}$

$= > p.e = - mg R$

∴ At A Distance (nR+R) from the Centre of the earth,

$p.e = \frac{-GMm}{n R +R } = \frac{-GMm}{(n + 1) R }$

As,

$GM = g { R}^{2}$

So,

$= > p.e = \frac{ - mg { R}^{2} }{(n + 1 ) R} = \frac{ - mg R }{ (n + 1)}$

∴ Increase in p.e =$\frac{ - mg R }{ (n + 1)} - (- mg R )$

$= > - mg R ( \frac{1}{n + 1} - \frac{n + 1 }{n + 1})$

$= > - mg R ( \frac{1 - n - 1}{n + 1} )$

$= > mg R ( \frac{n}{n + 1} )$

We Also Know that

Kinetic Energy= Increase In Potential Energy.

So,

$\frac{1}{2} m {v}^{2} = mg R ( \frac{n}{n + 1} )$

$= > \frac{1}{2} {v}^{2} = g R ( \frac{n}{n + 1} )$

$= > {v}^{2} = 2 \times g R ( \frac{n}{n + 1} )$

$= > v = \sqrt{2 \times g R ( \frac{n}{n + 1} )}$

As According To The Question

Options are in GM form

So We Will Substitute the Value of g

i.e,

$g = \frac{GM }{ {R}^{2} }$

$= > v = \sqrt{2 \times \frac{GM }{ {R}^{2} } \times R ( \frac{n}{n + 1} )}$

$= > v = \sqrt{2 \times \frac{GM }{ R } \times ( \frac{n}{n + 1} )}$

$= > v = \sqrt{2 ( \frac{n}{n + 1} ) \frac{GM }{ R } }$

Hence, Option C) Is Your Answer.