The mass of a planet and its diameter are thr | vedclass

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Question

(A) The acceleration due to gravity on the surface of the Earth is given by $g = \frac{GM_e}{R_e^2} = 9.8 \ m/s^2$.For the planet,the mass $M_p = 3M_e

Vedclass · Accepted Answer

$3.3$

Vedclass · Answer

(A) The acceleration due to gravity on the surface of the Earth is given by $g = \frac{GM_e}{R_e^2} = 9.8 \ m/s^2$.For the planet,the mass $M_p = 3M_e$ and the radius $R_p = 3R_e$ (since the diameter is three times that of the Earth,the radius is also three times).The acceleration due to gravity on the surface of the planet is $g_p = \frac{GM_p}{R_p^2}$.Substituting the values: $g_p = \frac{G(3M_e)}{(3R_e)^2} = \frac{3GM_e}{9R_e^2} = \frac{1}{3} \left( \frac{GM_e}{R_e^2} \right)$.Therefore,$g_p = \frac{g}{3} = \frac{9.8}{3} \approx 3.3 \ m/s^2$.

The mass of a planet and its diameter are three times those of the Earth. Then the acceleration due to gravity on the surface of the planet is ....... $m/s^2$.

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