Question

the kinetic energy of a body of mass M at a height h is equal to R by 2 from earth surface when mass M is thrown from surface with root GR​

Answer 1

ANSWER

Given h=

5

R

Potential Energy at surface of the earth, (PE

1

)=−

R

GMm

Potential Energy at height h=

5

R

, (PE

2

)=−

R+

5

R

GMm

=−

6R

5GMm

Increase in Potential energy=PE

2

−PE

1

=−

6R

5GMm

+

R

GMm

=

6R

GMm

(Using g=

R

2

GM

and h=

5

R

) ⇒PE

2

−PE

1

=

6

5

Answer 2

Answer:

Change in potential energy $=\frac{2}{3} \mathrm{mgR}$

Explanation:

Given: mass M at a height h is equal to R by 2 from earth surface when mass M is thrown from surface with root GR​

To find: Change in potential energy

Solution:

Mass $=m$, height $=2 R$

At height the acceleration due to gravity decreases, hence the value of acceleration due to gravity becomes$g^{\prime}=\frac{g}{1+h / R}=\frac{g R}{h+R}$

The value of potential energy $=m g h=\frac{m g h R}{h+R}$

Now $\mathrm{h}=2 \mathrm{R}$ (given)

Hence change in potential energy $=\frac{2}{3} \mathrm{mgR}$