Question:
The distance between the sun and the earth be $r$, then the angular momentum of the earth around the sun is proportional to
The distance between the sun and the earth be $r$, then the angular momentum of the earth around the sun is proportional to
Updated On: Jun 15, 2022
- $\sqrt{r}$
- $r^{3/2}$
- $r$
- None of these
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The Correct Option is A
Solution and Explanation
By Kepler's third law of planetary motion,
we have, $T^{2} \propto a^{3}$
$\Rightarrow T \propto a^{3 / 2}$
Given a $=r$ and $\omega=\frac{2 \pi}{T}$
= angular speed $=\omega \propto r^{-3 / 2}$
Also angular momentum $L=m r^{2} \omega$
$\Rightarrow L \propto r^{2} \times r^{-3 / 2}$
$\Rightarrow L \propto r^{3 / 2}$
we have, $T^{2} \propto a^{3}$
$\Rightarrow T \propto a^{3 / 2}$
Given a $=r$ and $\omega=\frac{2 \pi}{T}$
= angular speed $=\omega \propto r^{-3 / 2}$
Also angular momentum $L=m r^{2} \omega$
$\Rightarrow L \propto r^{2} \times r^{-3 / 2}$
$\Rightarrow L \propto r^{3 / 2}$
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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].