Question 297: Gravitation

Question

Assuming that the earth is revolving around the sun in circular orbit of radius R , the angular momentum is directly proportional to $\mathrm{R}^{\mathrm{n}}$. The value of ' $n$ ' is

Accepted Answer

The angular momentum of the Earth revolving around the Sun can be derived using the formula for angular momentum in circular motion:
$$ L = mvr $$
where:
$ L $ is the angular momentum,
$ m $ is the mass of the Earth,
$ v $ is the tangential velocity,
$ r $ is the radius of the circular orbit, given as $ R $.
For circular orbits under the gravitational influence of the Sun, the centripetal force is provided by gravitational force:
$$ \frac{mv^2}{r} = \frac{GMm}{r^2} $$
where:
$ M $ is the mass of the Sun,
$ G $ is the gravitational constant.
From this, solving for the velocity $ v $, we have:
$$ v^2 = \frac{GM}{r} $$
Taking the square root:
$$ v = \sqrt{\frac{GM}{r}} $$
Substituting this back into the angular momentum expression:
$$ L = m 	imes \sqrt{\frac{GM}{r}} 	imes r $$
Simplifying the equation:
$$ L = m 	imes r 	imes \sqrt{\frac{GM}{r}} $$
The term $ \sqrt{\frac{GM}{r}} $ simplifies to:
$$ L = m 	imes \sqrt{GMr} $$
Since the problem states that angular momentum $ L $ is directly proportional to $ R^n $, we can compare this with
$$ L \propto \sqrt{r} $$
or:
$$ L \propto r^{0.5} $$
Thus, $ n = 0.5 $. Therefore, the value of $ n $ is 0.5.
Correct Option: D. 0.5

Answer

2

Answer

1.5

Answer

1

Answer

0.5

Gravitation

Assuming that the earth is revolving around the sun in circular orbit of radius R , the angular momentum is directly proportional to \mathrm{R}^{\mathrm{n}}. The value of ' n ' is

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