Question

What is the kinetic energy of a body of mass m at a height h = $\frac{R}{2}$ from earth surface, when mass m is thrown from surface with $\sqrt{gR}$ (R is radius of earth)

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Answer 1

Solution :

Given :-

A body of mass m is thrown in upward direction from the surface of earth with velocity √(gR).

Where, R = radius of earth

To Find :-

Kinetic energy of body at height h equals to R/2 from the surface of earth.

Concept :-

This question is completely based on concept of Energy conservation.

Let, mass of earth = M

Calculation :-

$\implies\sf\red{Initial\:energy=Final\:energy}\\ \\ \implies\sf\:(PE)_1+(KE)_1=(PE)_2+(KE)_2\\ \\ \implies\sf\:-\dfrac{GMm}{R}+\dfrac{1}{2}m(\sqrt{gR})^2=-\dfrac{GMm}{R+\frac{R}{2}}+(KE)_2\\ \\ \implies\sf\:-\dfrac{GMm}{R}+\dfrac{mgR}{2}=-\dfrac{2GMm}{3R}+(KE)_2\\ \\ \implies\sf\:-\dfrac{GMm}{R}+\dfrac{GMm}{2R}=-\dfrac{2GMm}{3R}+(KE)_2\\ \\ \purple{\because\sf\:Gravitational\:acc.(g)=\dfrac{GM}{R^2}}\\ \\ \implies\sf\:-\dfrac{GMm}{R}+\dfrac{GMm}{2R}+\dfrac{2GMm}{3R}=(KE)_2\\ \\ \implies\sf\:\dfrac{-6GMm+3GMm+4GMm}{6R}=(KE)_2\\ \\ \implies\sf\:\dfrac{GMm}{6R}=(KE)_2\\ \\ \implies\underline{\boxed{\bf{\pink{(KE)_2=\dfrac{mgR}{6}=\dfrac{GMm}{6R}}}}}$

Answer 2

According to the question;

The body of mass m at a height h = $\frac{R}{2}$ from the earth's surface

When mass m is thrown from the surface with $\sqrt{gR}$

where R = radius of the earth.

now,

Initial energy = final energy

(P.E)₁ + (K.E)₁ = (P.E)₂ + (K.E)₂

$-\frac{GMm}{R} + \frac{1}{2} m (\sqrt{gR})^{2} = -\frac{GMm}{R+\frac{R}{2} } + (K.E)_{2} \\-\frac{GMm}{R} + \frac{mgR}{2} = -\frac{GMm}{3R} + (K.E)_{2}\\-\frac{GMm}{R} + \frac{GMm}{2R} = -\frac{GMm}{3R} + (K.E)_{2}\\Gravitational acc.(g) = \frac{GM}{R^{2} } \\ (K.E)_{2} = -\frac{GMm}{R} + \frac{GMm}{2R} + \frac{GMm}{3R}\\ (K.E)_{2} = \frac{- 6GMm + 3GMm + 4GMm}{6R} \\ (K.E)_{2} = \frac{GMm}{6R}$

The kinetic energy of a body of mass m at a height h = $\frac{R}{2}$ from the earth's surface, when mass m is thrown from surface with $\sqrt{gR}$ is $\frac{GMm}{6R}$ .

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