Question:

A saturn year is $29.5$ times the earth year. How far is the saturn from the sun if the earth is $1.5 \times 10^8\, km$ away from the sun?

Updated On: Jul 6, 2022

$1.4 \times 10^6\, km$
$1.4 \times 10^7\, km$
$1.4 \times 10^8\, km$
$1.4 \times 10^9\, km$

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The Correct Option is D

Solution and Explanation

Here, $T_S = 29.5T_E$, $R_E= 1.5 \times 10^8\, km$, $R_S = ?$ According to Keplers third law, $T^{2} \propto R^{3}$ $\therefore \frac{T^{2}_{S}}{T^{2}_{E}} = \frac{R^{3}_{S}}{R^{3}_{E}}$; $R_{S} = R_{E} \left(\frac{T_{S}}{T_{E}}\right)^{2/3}$ $R_{S} = 1.5 \times 10^{8}\left(\frac{29.5T_{E}}{T_{E}}\right)^{2/3}$ $= 1.4 \times 10^{9}\,km$

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Notes on Gravitation

Concepts Used:

Gravitation

In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.

Newton’s Law of Gravitation

According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,

F ∝ (M₁M₂) . . . . (1)
(F ∝ 1/r²) . . . . (2)

On combining equations (1) and (2) we get,

F ∝ M₁M₂/r²

F = G × [M₁M₂]/r² . . . . (7)

Or, f(r) = GM₁M₂/r²

The dimension formula of G is [M^-1L³T^-2].