JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 11)
Consider two satellites S1 and S2 with periods of revolution 1 hr. and 8 hr. respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S1 to the angular velocity of satellite S2 is :
1 : 4
8 : 1
2 : 1
1 : 8
Explanation
Given, period of revolution of first satellite,
T1 = 1h
Period of revolution of second satellite,
T2 = 8h
$$\therefore$$ $${{{T_1}} \over {{T_2}}} = {1 \over 8}$$
We know that, $$\omega = {{2\pi } \over T}$$
$$ \Rightarrow \omega \propto {1 \over T}$$
$$\because$$ $${{{\omega _1}} \over {{\omega _2}}} = {{{T_2}} \over {{T_1}}} \Rightarrow {{{\omega _1}} \over {{\omega _2}}} = {8 \over 1}$$
or $${\omega _1}:{\omega _2} = 8:1$$
T1 = 1h
Period of revolution of second satellite,
T2 = 8h
$$\therefore$$ $${{{T_1}} \over {{T_2}}} = {1 \over 8}$$
We know that, $$\omega = {{2\pi } \over T}$$
$$ \Rightarrow \omega \propto {1 \over T}$$
$$\because$$ $${{{\omega _1}} \over {{\omega _2}}} = {{{T_2}} \over {{T_1}}} \Rightarrow {{{\omega _1}} \over {{\omega _2}}} = {8 \over 1}$$
or $${\omega _1}:{\omega _2} = 8:1$$
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