91Ó°ÊÓ

The given data can be listed below as,

• The mass of the hammer is, m.
• Theradius of the Earth, R_E.
• The mass of the Earth is, m_E.
• The speed the hammer is, v.
• The height of hammer from Earth’s surface is, h.

According to law of conservation of energy, the total energy of a system remains constant irrespective of the position of the system. Here, total energy refers to the sum of all the energies like kinetic energy, potential energy etc.

State 1 in the figure shows the initial position, whereas the state 2 shows the final position of the hammer.

The distance r₁ can be written as,

$r_1={\text{R}}_{\text{E}}+h$

Here, is the distance from the Earth’s centre to point 1.

The distance r₂ will be,

$r_2={\text{R}}_{\text{E}}$

Here, is the distance from the Earth’s centre to point 2.

Apply the conservation of the energy at point 1 and 2. It can be expressed as,

$K_1+U_1+W_o=K_2+U_2$ …... (I)

Here, K_1, K₂, and U_1,U₂are the kinetic gravitational potential energy, and W_o is the external work.

Only gravity does work. So, external work (W_o ) will be zero, i.e., W_o = 0.

At point 1,

The kinetic energy at point 1 is expressed as,

K₁= 0

Because initial velocity is zero.

The gravitational potential energy at state 1 is expressed as,

$U_1=−G\frac{m{\text{m}}_{\text{E}}}{r_1}$

Here, G is the gravitational constant.

Substitute R_E + h for in the above equation.

$U_1=−G\frac{m{\text{m}}_{\text{E}}}{{\text{R}}_{\text{E}}+h}$

At point 2,

The kinetic energy is expressed as,

$K_2=\frac{1}{2}m{v}_2^2$

Here, v₂ is the speed of hammer at point 2.

The gravitational potential energy at state 2 is expressed as,

$U_2=−G\frac{m{\text{m}}_{\text{E}}}{r_2}$

Substitute R_E for r₂in the above equation.

$U_2=−G\frac{m{\text{m}}_{\text{E}}}{{\text{R}}_{\text{E}}}$

Substitute 0 for K₁, $−G\frac{m{\text{m}}_{\text{E}}}{{\text{R}}_{\text{E}}+h}$ for U₁, $\frac{1}{2}m{v}_2^2$ for K₂, and $−G\frac{m{\text{m}}_{\text{E}}}{{\text{R}}_{\text{E}}}$ for U₂in equation (I)

$\begin{array}{c}0−G\frac{m{\text{m}}_{\text{E}}}{{\text{R}}_{\text{E}}+h}=\frac{1}{2}m{v}_2^2−G\frac{m{\text{m}}_{\text{E}}}{{\text{R}}_{\text{E}}}\\ v_2^2=2G{\text{m}}_{\text{E}}\left(\frac{1}{{\text{R}}_{\text{E}}}−\frac{1}{{\text{R}}_{\text{E}}+h}\right)\\ v_2^2=2G{\text{m}}_{\text{E}}\left(\frac{{\text{R}}_{\text{E}}+h−{\text{R}}_{\text{E}}}{{\text{R}}_{\text{E}}\left({\text{R}}_{\text{E}}+h\right)}\right)\\ v_2^2=\frac{2G{\text{m}}_{\text{E}}h}{{\text{R}}_{\text{E}}\left({\text{R}}_{\text{E}}+h\right)}\\ v_2=\sqrt{\frac{2G{\text{m}}_{\text{E}}h}{{\text{R}}_{\text{E}}\left({\text{R}}_{\text{E}}+h\right)}}\end{array}$

Hence, the speed of the hammer when it reaches the Earth’s surface is,$\sqrt{\frac{2G{\text{m}}_{\text{E}}h}{{\text{R}}_{\text{E}}\left({\text{R}}_{\text{E}}+h\right)}}$