JEE MAIN - Physics (2019 - 12th January Evening Slot - No. 20)
Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, TA/TB, is ;
2
$${{1 \over 2}}$$
$$\sqrt {{1 \over 2}} $$
1
Explanation
Orbital velocity V = $$\sqrt {{{GMe} \over r}} $$
TA = $${1 \over 2}$$ mA V$$_A^2$$
TB = $${1 \over 2}$$ mB V$$_B^2$$
$$ \Rightarrow $$ $${{{T_A}} \over {{T_B}}} = {{m \times {{Gm} \over R}} \over {2m \times {{Gm} \over {2R}}}}$$
$$ \Rightarrow $$ $${{{T_A}} \over {{T_B}}}$$ = 1
TA = $${1 \over 2}$$ mA V$$_A^2$$
TB = $${1 \over 2}$$ mB V$$_B^2$$
$$ \Rightarrow $$ $${{{T_A}} \over {{T_B}}} = {{m \times {{Gm} \over R}} \over {2m \times {{Gm} \over {2R}}}}$$
$$ \Rightarrow $$ $${{{T_A}} \over {{T_B}}}$$ = 1
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