The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is 9:4. The mass of the planet is $\frac{1}{9}th$ 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)

Option 1)
$\frac{R}{3}$

Option 2)
$\frac{R}{4}$

Option 3)
$\frac{R}{9}$

Option 4)
$\frac{R}{2}$

Answers (1)

S solutionqc

Newton's Law of Gravitation -

$F\; \alpha\; \frac{m_{1}m_{2}}{r^{2}}$

$F\; = \frac{G\, m_{1}\, m_{2}}{r^{2}}$

$F\rightarrow$ Force

$G\rightarrow$ Gravitalional constant

$m_1,m_2\rightarrow$ Masses

$r\rightarrow$ Distance between masses

- wherein

Force is along the line joining the two masses

1 - planet 2- Earth

$W_{1}=\frac{GM_{1}m}{R^{2}_{1}}$ $W_{2}=\frac{GM_{2}m}{R^{2}_{2}}$

$\frac{W_{1}}{W_{2}}=\frac{M_{1}}{M_{2}}\left ( \frac{R_{2}^{2}}{R_{1}^{2}} \right )$

$\frac{4}{9}=\frac{1}{9}\left ( \frac{R_{2}^{2}}{R_{1}^{2}} \right )=\frac{R_{1}}{R_{2}}=\frac{1}{2}$

$Rplanet = \frac{Rearth}{2}$

Option 1)

$\frac{R}{3}$

Option 2)

$\frac{R}{4}$

Option 3)

$\frac{R}{9}$

Option 4)

$\frac{R}{2}$

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