The ratio of the weights of a body on Earth’s surface to that on the surface of a planet is 9:4 The mass of the planet is 1/9th 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)

Correct answer: (B) R/2. Weight =mGM/(R^2) ∴9/4=GM_e/Re2×R/p2GM_p Given that, M_p=M_e/9 ∴9/4=M_e/M_e9×R/p2Re2 Rp2=Re2×1/4 R_p=R_e/2 As, R_e=R ⇒R_p=R/2

Q78 / 90 · JEE Main 2019 (12 Apr Shift 2) · Physics

The ratio of the weights of a body on Earth’s surface to that on the surface of a planet is $9 : 4$ The mass of the planet is $\frac{1}{9} t h$ 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)

A
$\frac{R}{4}$
Correct answer
B
$\frac{R}{2}$
C
$\frac{R}{3}$
D
$\frac{R}{9}$

Solution

Weight $= m (\frac{G M}{R^{2}})$
$∴ \frac{9}{4} = \frac{G M_{e}}{R_{e}^{2}} \times \frac{R_{p}^{2}}{G M_{p}}$
Given that, $M_{p} = \frac{M_{e}}{9}$
$∴ \frac{9}{4} = \frac{M_{e}}{(\frac{M_{e}}{9})} \times \frac{R_{p}^{2}}{R_{e}^{2}}$
$R_{p}^{2} = R_{e}^{2} \times \frac{1}{4}$
$R_{p} = \frac{R_{e}}{2}$
As, $R_{e} = R$
$\Rightarrow R_{p} = \frac{R}{2}$

Hence, the correct answer is option: 'B'.

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Solution

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The ratio of the weights of a body on Earth’s surface to that on the surface of a planet is $9 : 4$ The mass of the planet is $\frac{1}{9} t h$ 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)