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Question
A small block of superdense material has a mass of 3 × 1024kg. It is situated at a height h (much smaller than the earth's radius) from where it falls on the earth's surface. Find its speed when its height from the earth's surface has reduce to to h/2. The mass of the earth is 6 × 1024kg.
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Solution
It is given that h is much lesser than the radius of the earth.
Mass of the earth, Me = 6 × 1024 kg
Mass of the block, Mb = 3 × 1024 kg
Let Ve be the velocity of the earth and Vb be the velocity of the block.
Let the earth and the block be attracted by gravitational force.
Thus, according to the conservation law of energy, the change in gravitational potential energy will be the K.E. of the block.
`GM_eM_b(1/(R + (h/2)) - 1/(R + h)) = (1/2)M_e xx V_e^2 + (1/2)M_b xx V_b^2` ...(i)
The momentum is conserved as only the internal force acts in this system.
MeVe = MbVb
⇒ `V_e = (M_bV_b)/M_e` ...(ii)
Putting in equation (i),
`GM_eM_b[2/(2R + h) - 1/(R + h)]`
= `(1/2) xx M_e^2 xx (M_b^2V_b^2)/M_e^2 + (1/2) xx M_b xx V_b^2`
= `(1/2) xx V_b^2 M_b/M_e + 1/2 xx M_b xx V_b^2`
⇒ `GM_e (2r + 2h - 2R - h)/((2R + h)(R + h))`
= `(1/2) xx V_b^2 xx ((3 xx 10^24)/(6 xx 10^24) + 1)`
⇒ `[(GM xx h)/(2R^2)] = (1/2) xx V_b^2 xx (3/2)`
⇒ `gh = V_b^2 xx (3/2)`
`V_b = (2gh)/3`
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