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# A body of mass m rises to height$h=\frac{R}{5}$from the earth's surface, where R is the earth's radius. If g is the acceleration due to gravity at the earth's surface, the increase in potential energy is

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## Step 1: Given data The radius of earth is R.Mass of the body is m.The height of the body from the earth's surface is $h=\frac{R}{5}$ .Step 2: Gravitational potential energyThe gravitational potential energy at any point in a gravitational field of a system of masses is the amount of work done in bringing a unit mass from infinity to that point.Gravitational potential energy due to a point mass m at a distance r is, $V=-{\int }_{\infty }^{r}GM\frac{m}{{x}^{2}}dx=-GM\frac{m}{r}$ , where G is the universal gravitational constant and M is the mass of the earth.Step 3: Finding the change in gravitational potential energyWe know that the gravitational potential energy of a body of mass m placed on the earth's surface is$V=-G\frac{Mm}{R}$ ……………(1)where M and R are the mass of the earth and the radius of the earth.Now, gravitational potential energy at a height $h=\frac{R}{5}$ from the earth's surface is, $V\text{'}=-G\frac{Mm}{R+\frac{R}{5}}=-G\frac{5Mm}{6R}\phantom{\rule{0ex}{0ex}}orV\text{'}=-G\frac{5Mm}{6R}..................\left(2\right)$So, the change in gravitational potential energy is $∆V=V\text{'}-V=-G\frac{5mM}{6R}-\left(-G\frac{mM}{R}\right).\phantom{\rule{0ex}{0ex}}or∆V=-G\frac{5mM}{6R}+G\frac{mM}{R}=G\frac{mM}{6R}.\phantom{\rule{0ex}{0ex}}or∆V=G\frac{mM}{6R}.$(From equations 1 and 2 )Therefore, the increase in gravitational potential energy is $G\frac{Mm}{6R}$.

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