Question 60: Gravitation

Question

Gravitational acceleration on the surface of a planet is $${{\sqrt 6 } \over {11}}g$$, where $$g$$ is the gravitational acceleration on
the surface of the earth. The average mass density of the planet is $${2 \over 3}$$
times that of the earth. If the escape
speed on the surface of the earth is taken to be 11 kms-1, the escape speed on the surface of the planet in
kms-1 will be

Accepted Answer

On the planet,
$${g_p} = {{G{M_p}} \over {R_p^2}} = {G \over {R_p^2}}\left( {{4 \over 3}\pi R_p^3{ho _p}} ight) = {4 \over 3}G\pi R_p^{}{ho _p}$$
On the earth,
$${g_e} = {{G{M_e}} \over {R_e^2}} = {G \over {R_e^2}}\left( {{4 \over 3}\pi R_e^3{ho _e}} ight) = {4 \over 3}G\pi R_e^{}{ho _e}$$
$$	herefore$$ $${{{g_p}} \over {{g_e}}} = {{R_p^{}{ho _p}} \over {R_e^{}{ho _e}}}$$ or $${{{R_p}} \over {{R_e}}} = {{g_p^{}{ho _p}} \over {g_e^{}{ho _e}}}$$ ..... (i)
On the planet, $${v_p} = \sqrt {2{g_p}{R_p}} $$
On the earth, $${v_e} = \sqrt {2{g_e}{R_e}} $$
$$	herefore$$ $${{{v_p}} \over {{v_e}}} = \sqrt {{{{g_p}{R_p}} \over {{g_e}{R_e}}}}  = {{{g_p}} \over {{g_e}}}\sqrt {{{{ho _e}} \over {{ho _p}}}} $$ (Using (i))
Here, $${ho _p} = {2 \over 3}{ho _e}$$, $${g_p} = {{\sqrt 6 } \over {11}}{g_e}$$
$$	herefore$$ $${{{v_p}} \over {{v_e}}} = {{\sqrt 6 } \over {11}}\sqrt {{3 \over 2}} $$
or, $${v_p} = 11 	imes {{\sqrt 6 } \over {11}} 	imes \sqrt {{3 \over 2}} $$ ($$\because$$ v
e
= 11 km s
$$-$$1
(Given))
= 3 km s
$$-$$1

Gravitation

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