Physics, asked by vishwas53, 11 months ago

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

Answers

Answered by joshikapatnaik007
4

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

mv^{2}/2 = gRm/5

v = \sqrt{2gR/5}

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Answered by KaurSukhvir
0

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.

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