First slide

Motion of planets and satellites

Question

In the following four periods
(i) Time of revolution of a satellite just above the earth’s surface $(T_{s t})$
(ii) Period of oscillation of mass inside the tunnel bored along the diameter of the earth $(T_{m a})$
(iii) Period of simple pendulum having a length equal to the earth’s radius in a uniform field of 9.8 N/kg $(T_{s p})$
(iv) Period of an infinite length simple pendulum in the earth’s real gravitational field $(T_{i s})$

Easy

A

$T_{s t} > T_{m a}$
B

$T_{m a} > T_{s t}$
C

$T_{s p} < T_{i s}$
D

$T_{s t} = T_{m a} = T_{s p} = T_{i s}$

Solution

$(i) T_{s t} = 2 π \sqrt{\frac{{(R + h)}^{3}}{G M}} = 2 π \sqrt{\frac{{(R + 0)}^{3}}{g R^{2}}} = 2 π \sqrt{\frac{R}{g}}$
$[A s h < < R a n d G M = g R^{2}] (i i) T_{m a} = 2 π \sqrt{\frac{R}{g}} (i i i) T_{s p} = 2 π \sqrt{\frac{1}{g (\frac{1}{l} + \frac{1}{R})}} = 2 π \sqrt{\frac{1}{g (\frac{1}{R} + \frac{1}{R})}} 2 π \sqrt{\frac{1}{g (\frac{2}{R})}} = = 2 π \sqrt{\frac{R}{2 g}} [A s l = R] (i v) T_{i s} = 2 π \sqrt{\frac{1}{g (\frac{1}{l} + \frac{1}{R})}} = 2 π \sqrt{\frac{1}{g (\frac{1}{\infty} + \frac{1}{R})}} = 2 π \sqrt{\frac{R}{g}} [A s l = \infty] F r o m (i i i) a n d (i v) T_{s p} < T_{i s}$

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