# Q. The energy required to take a satellite to a height 'h' above Earth surface (radius of Earth = $6.4 \times 10^3 \; km)$ is $E_1$ and kinetic energy required for the satellite to be in a circular orbit at this height is $E_2$. The value of h for which $E_1$ and $E_2$ are equal, is:

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Solution:

## $U_{surface} + E_1 = U_h$ KE of satelite is zero at earth surface & at height h $- \frac{GM_{e}m}{R_{e}} + E_{1} = - \frac{GM_{e}m}{\left(R_{e}+h\right)}$ $E_{1} =GM_{e}m \left( \frac{1}{R_{e}} - \frac{1}{R_{e} +h}\right)$ $E_{1} = \frac{GM_{e}m}{\left(R_{e} +h\right)} \times\frac{h}{R_{e}}$ Gravitational attraction $F_{G} = ma_{C} = \frac{mv^{2}}{\left(R_{e} +h\right)}$ $E_{2 } \Rightarrow \frac{mv^{2}}{\left(R_{e} +h\right)} = \frac{GM_{e}m}{\left(R_{e} + h\right)^{2}}$ $mv^{2} = \frac{GM_{e}m}{\left(R_{e} +h\right) }$ $E_{2} = \frac{mv^{2}}{2} = \frac{GM_{e}m}{2 \left(R_{e} + h\right)}$ $E_{1} =E_{2}$ $\frac{h}{R_{e}} = \frac{1}{2} \Rightarrow h = \frac{R_{e}}{2} = 3200$ km

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