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+2 votes
39.6k views
in Physics by (30.6k points)
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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) R\((\frac{k}{1-k})^2\)

(2) R\((\frac{k}{1+k})^2\)

(3) \(\frac{R^2k}{1+k}\)

(4) \(\frac{Rk^2}{1-k^2}\)

4 Answers

+3 votes
by (1.4k points)
selected by
 
Best answer

We know that,

escape velocity ve\(\sqrt{\frac{2GM}{R}}\) 

conservation of energy

Ei = Ef = 0

Ei - Ef = 0

Ei \(\frac{1}{2}\)m(kve)2 - \(\frac{GMm}{R}\) .........(1)

Ef\(-\frac{GMm}{R}\) ............(2)

Equal equation (1) and (2)

\(\frac{1}{2}\)m(k)2\(\frac{2GM}{R}\)\(\frac{GMm}{R}\) = \(-\frac{GMm}{R+h}\)

k2\(\frac{GMm}{R}\) - \(\frac{GMm}{R}\) = \(-\frac{GMm}{R+h}\)

k2\(\frac{GMm}{R}\) = \(\frac{GMm}{R}\) \(-\frac{GMm}{R+h}\) 

k2\(\frac{GMm}{R}\) = GMm (\(\frac{1}{R}\) - \(\frac{1}{R+h}\))

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

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

k2\(\frac{h}{R+h}\)

\(\frac{R+h}{h}\) = \(\frac{1}{k^2}\)

⇒ \(\frac{R}{h}\) + 1 = \(\frac{1}{k^2}\)

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

h = \(\frac{R\,k^2}{1-k^2}\)

+4 votes
by (30.7k points)

Option : (4). \(\frac{R^2k}{1-k^2}\)

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

+1 vote
by (115 points)
edited by

Using energy conservation

+1 vote
by (17.0k points)

Correct option is (4) \(\frac{Rk^2}{1 - k^2}\)

From conservation of mechanical energy,

\(\frac 12mv^2 - \frac{GmM}R = - \frac{GmM}{(R+ h)}\)

⇒ \(\frac{v^2}2 = \frac{GM}R - \frac{GM}{(R + h)}\)

\(= \frac{GMh}{R(R + h)}\)

⇒ \(\frac 12 k^2 v^2_e= \frac{GMh}{R(R + h)}\)

We know that, 

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

\(\frac 12 k^2 (\frac {2GM}R) = \frac{GMh}{R(R + h)}\)

⇒ \(k^2 = \frac h{R + h}\)

⇒ \(Rk^2 + hk^2 = h\)

⇒ \(Rk^2 = h(1 - k^2)\)

⇒ \(h = \frac{Rk^2}{1 - k^2}\)

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