Question

A body of mass m is raised to a height h above the surface of the earth of mass m and radius r until its gravitational potential energy increases by 1/3 mgr .the value of h is?

Answer 1

Answer:

h = R/2

Explanation:

R is radius of the earth

G is universal gravitation constant

M is the mass of the earth

Answer 2

The value of h is $\dfrac{r}{2}$

Explanation:

Given that,

Mass of body = m

Height above the surface= h

Radius of earth = r

Mass of earth = m

Change potential energy $\Delta U=\dfrac{1}{3}mgr$

Potential energy on surface of earth is,

$U_{e}=-\dfrac{GmM}{r}$

Potential energy at height h is,

$U_{h}=-\dfrac{GmM}{(r+h)}$

We need to calculate the value of h

Using formula of of change in potential energy

$\Delta U=U_{h}-U_{r}$

Put the value into the formula

$\dfrac{GmM}{3r}=-\dfrac{GmM}{(r+h)}+\dfrac{GmM}{r}$

$\dfrac{1}{3r}=-\dfrac{1}{r+h}+\dfrac{1}{r}$

$\dfrac{1}{3r}=\dfrac{-r+r+h}{r(r+h)}$

$r^2+rh=3rh$

$h=\dfrac{r}{2}$

Hence, The value of h is $\dfrac{r}{2}$

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Topic : potential energy

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