Question #61455

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Question #61455

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Answer 1 · 2016-10-19T23:49:20

(b) $\frac{1}{3}$ $1/3$

Explanation:

As per Law of Universal Gravitation the force of attraction between two bodies is given as

$F_{G} = G \frac{M_{1} . M_{2}}{r^{2}}$ $F_G =G (M_1.M_2)/r^2$

Where $G$ $G$ is gravitational constant.
It has the value $6.67408 \times 10^{- 11} m^{3} k g^{- 1} s^{- 2}$ $6.67408 xx 10^-11 m^3 kg^-1 s^-2$ .
Comparing it with the expression
$F (weight) = m g$ $F("weight")=mg$
we get value of $g = G \frac{M_{e}}{r_{e}^{2}}$ $g=G (M_e)/r_e^2$ ......(1)
If $ρ$ $rho$ is density of earth, then
$ρ = \frac{M_{e}}{V o l u m e}$ $rho=M_e/(Volume)$
$ρ = \frac{M_{e}}{\frac{4}{3} π r_{e}^{3}}$ $rho=M_e/(4/3pir_e^3)$
$\Rightarrow M_{e} = \frac{4}{3} π ρ r_{e}^{3}$ $=>M_e=4/3pirhor_e^3$
Inserting above value of mass of earth in (1) we get
$g = G \frac{\frac{4}{3} π ρ r_{e}^{3}}{r_{e}^{2}}$ $g=G (4/3pirhor_e^3)/r_e^2$
$\Rightarrow g = G \frac{4}{3} π ρ r_{e}$ $=>g=G 4/3pirhor_e$
$\Rightarrow g \propto ρ r_{e}$ $=>gproprhor_e$
From above we see that if radius of earth is increased by a factor of $3$ $3$ , to keep value of $g$ $g$ same we need to decrease value of $ρ$ $rho$ by a factor of $3$ $3$ .
Hence we get $\frac{1}{3}$ $1/3$

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Question #61455

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