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7.Gravitation
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Acceleration due to gravity at surface of a planet is equal to that at surface of earth and density is $1.5$ times that of earth. If radius of earth is $R$, radius of planet is .................
A
$\frac{3}{2} R$
B
$\frac{2}{3} R$
C
$\frac{9}{4} R$
D
$\frac{4}{9} R$
Solution
(b)
Gravity at surface of a planet $=$ gravity at surface of the earth.
$g_p=g_e$
$P_1=1.5 P_e$
Radius of earth $=R$
$g=\frac{G\left(P \times \frac{4}{3} \pi R^3\right)}{R^2}$
It's given, $P_P=1.5 P_E$
$\frac{ G \times \frac{4}{3} \pi R^3 \times P_e}{R^2}=\frac{G \times \frac{4}{3} \pi \times\left(R^1\right)^3 \times P_p}{\left(R^1\right)^2}$
or, $R_P=R^1 P_P$
or, $R^1=R \times \frac{P_p}{P_p}=\frac{R}{1.5}=\frac{2 R}{3}$.
Standard 11
Physics
