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 1/9 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) (a) R/9 (b) R/2 (c) R/3 (d) R/4

Answer: (b). g = GM/R² ⇒ (gp)/(ge) = (Mp)/(Me)(R/Rp)² 4/9 = 1/9(R/Rp)² ⇒ (R/Rp)² = 4 So the answer is (b) Rp = R/2.

The ratio of the weights of a body on the Earth’s surface to that on… | JEE Main 2019 (12 Apr, Shift 2)

Q112JEE Main 2019 · 12 APR · SHIFT 2Medium

Acceleration Due to Gravity

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 $\frac{1}{9}$ 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)

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Answer: (b)

$g = \frac{GM}{R^2} \Rightarrow \frac{g_p}{g_e} = \frac{M_p}{M_e}\left(\frac{R}{R_p}\right)^2$
$\frac{4}{9} = \frac{1}{9}\left(\frac{R}{R_p}\right)^2 \Rightarrow \left(\frac{R}{R_p}\right)^2 = 4$
So the answer is (b) $R_p = \frac{R}{2}$ .

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Acceleration Due to Gravity question from JEE Main 2019 (12 Apr, Shift 2)

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