Question 202: Gravitation

Question

The ratio of the weights of a body on the Earth’s surface to that on the surface of a planets is 9 : 4. The mass
of the planet is
$${1 \over 9}$$
th of that of the Earth. If 'R' is the radius of the Earth, what is the radius of the planet ?
(Take the planets to have the same mass density)

Accepted Answer

W = mg
as m = constant everywhere
$$ 	herefore $$ W $$ \propto $$ g
$${{{g_E}} \over {{g_p}}}$$ = $${9 \over 4}$$
We know,
$$g = {{GM} \over {{R^2}}}$$
$$ 	herefore $$ $${{{g_E}} \over {{g_p}}} = {{{M_E}} \over {{M_p}}} 	imes {{R_p^2} \over {R_E^2}}$$
$$ \Rightarrow $$$${9 \over 4} = {9 \over 1} 	imes {{R_p^2} \over {R_E^2}}$$
$$ \Rightarrow $$ $${{{R_p}} \over {{R_E}}} = {1 \over 2}$$
$${R_p} = {{{R_E}} \over 2}$$ = $${R \over 2}$$ [Here R
E
= R ]

Answer

$${R \over 9}$$

Answer

$${R \over 2}$$

Answer

$${R \over 3}$$

Answer

$${R \over 4}$$

Gravitation

The ratio of the weights of a body on the Earth’s surface to that on the surface of a planets is 9 : 4. The mass of the planet is ${1 \over 9}$ th of that of the Earth. If 'R' is the radius of the Earth, what is the radius of the planet ? (Take the planets to have the same mass density)

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