Test Index

Gravitation Part 1

Section: Physics

© examsnet.com

Question : 91 of 100

Marks: +1, -0

If

g

is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass

m

raised from the surface of the earth to a height equal to the radius

R

of the earth is

$\frac{1}{4} m g R$
$2 m g R$
$\frac{1}{2} m g R$
$m g R$

Solution:

Gravitational potential energy on the earth surface of a body

U = -\;\frac{G m M}{R}

And at the height

h

from the earth surface the potential energy

U_h = -\;\frac{G m M}{R+h} = -\;\frac{G m M}{2R}

[as

h = R

]

So the gain in the potential energy

\Delta U = U_h - U

\; \therefore \Delta U = \;\frac{-G m M}{2R} + \;\frac{G m M}{R}

\Rightarrow \Delta U = \;\frac{G m M}{2R}

Now

\frac{G M}{R^2} = g ; \;\;\therefore \;\frac{G M}{R} = g R

\therefore \Delta U = \frac{1}{2} m g R

© examsnet.com

Go to Question:

Prev Question

Next Question