Question

the minimum speed with which a body must be thrown to reach a height of R/4 above the earth's surface is? ​

Answer 1

Answer:

Explanation:

F=GMm/$x^{2}$

dW=Fdx

dW=GMmdx/$x^{2}$

integrating with limits R to 5R/4

W=∫ GMmdx/$x^{2}$

W= GMm ∫ 1/$x^{2}$ dx

W= GMm/5R

W= gRm/5

m$v^{2}$/2 = gRm/5

v = $\sqrt{2gR/5}$

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Answer 2

Answer:

A body is thrown with minimum speed equal to  $\sqrt{\frac{2GM}{5R} }$  to reach a height of $R/4$.

Explanation:

Given a body  is thrown at height $R/4$  from the surface of earth, where R is the radius of earth.

Final speed of body, $v=0$

From the law of conservation of energy, $E_{1}=E_{2}$             .............(1)

$E=K.E.+P.E.=\frac{1}{2}mv^{2}+[\frac{-GMm}{r}]$

Therefore, $E_{1}=\frac{1}{2}mv^{2}+[\frac{-GMm}{R}]$

$E_{2}=0+[\frac{-GMm}{R+\frac{R}{4} }]=[\frac{-GMm}{\frac{5R}{4} }]=\frac{-4GMm}{{5R} }$

Put the value of E₁ and E₂ in eq.(1),

$\frac{1}{2}mv^{2}+[\frac{-GMm}{R}]=\frac{-4GMm}{{5R} }$

$\frac{1}{2}v^{2} =\frac{-4GM}{5R}+\frac{GM}{R}$

$v=\sqrt{\frac{2GM}{5R} }$            

where $G$ is universal gravitational constant

and $M$ is the mass of earth.