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Q. A satellite is revolving in a circular orbit at a height $h$ above the surface of the earth of radius $R$ . The speed of the satellite in its orbit is one fourth the escape velocity from the surface of the earth. The relation between $h$ and $R$ is

NTA AbhyasNTA Abhyas 2020Gravitation

Solution:

$\upsilon_{0} = \frac{1}{4} \upsilon_{e}$
$\sqrt{\frac{G M}{\left(\right. R + h \left.\right)}} = \frac{1}{4} \sqrt{\frac{2 G M}{R}}$
$\frac{G M}{\left(\right. R + h \left.\right)} = \frac{1}{16} \times \frac{2 G M}{R}$
$R + h = 8 \left(\right. R \left.\right)$
$R + h = 8 R$
$7 R = h$