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An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the magnitude of the escape velocity from the earth. The height $(h)$ of the satellite above the earth's surface is (Take radius of earth as $R_{e}$ )
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Verified Answer
The correct answer is:
$h=R_{e}$
The escape velocity from earth is given by $v_{e}=\sqrt{2 \mathrm{~g} R_{e}} \ldots$ (i)
The orbital velocity of a satellite revolving around earth is given by
$$
v_{0}=\frac{\sqrt{\mathrm{G} M_{e}}}{\left(R_{e}+h\right)}
$$
where, $M_{e}=$ mass of earth, $R_{e}=$ radius of earth, $h=$ height of satellite from surface of earth.
By the relation $\mathrm{G} M_{e}=\mathrm{g} R_{e}^{2}$
$$
\text { So, } \quad v_{0}=\frac{\sqrt{g R_{e}^{2}}}{\left(R_{e}+h\right)} \ldots
$$
Dividing equation (i) by (ii), we get
$$
\frac{v_{e}}{v_{0}}=\frac{\sqrt{2\left(R_{e}+h\right)}}{\left(R_{e}\right)}
$$
Given, $v_{0}=\frac{v_{e}}{2}$
$$
\frac{2 v_{e}}{v_{e}}=\frac{\sqrt{2\left(R_{e}+h\right)}}{R_{e}}
$$
Squaring on both side, we get
$$
4=\frac{2\left(R_{e}+h\right)}{R_{e}}
$$
or $R_{e}+h=2 R_{e} \quad$ i.e., $h=R_{e}$
The orbital velocity of a satellite revolving around earth is given by
$$
v_{0}=\frac{\sqrt{\mathrm{G} M_{e}}}{\left(R_{e}+h\right)}
$$
where, $M_{e}=$ mass of earth, $R_{e}=$ radius of earth, $h=$ height of satellite from surface of earth.
By the relation $\mathrm{G} M_{e}=\mathrm{g} R_{e}^{2}$
$$
\text { So, } \quad v_{0}=\frac{\sqrt{g R_{e}^{2}}}{\left(R_{e}+h\right)} \ldots
$$
Dividing equation (i) by (ii), we get
$$
\frac{v_{e}}{v_{0}}=\frac{\sqrt{2\left(R_{e}+h\right)}}{\left(R_{e}\right)}
$$
Given, $v_{0}=\frac{v_{e}}{2}$
$$
\frac{2 v_{e}}{v_{e}}=\frac{\sqrt{2\left(R_{e}+h\right)}}{R_{e}}
$$
Squaring on both side, we get
$$
4=\frac{2\left(R_{e}+h\right)}{R_{e}}
$$
or $R_{e}+h=2 R_{e} \quad$ i.e., $h=R_{e}$
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