Class 11th Overview Physics Physics Gravitation

Suppose the gravitational force varies inversely as the $n^{t h}$ power of distance. Then, time period of a planet in circular orbit of radius R around the sum will be proportional to

Option 1 - <p><span class="mathml" contenteditable="false"> <math> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> </msup> </math> </span></p>

Option 2 - <p><span class="mathml" contenteditable="false"> <math> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>n</mi> <mo>-</mo> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </msup> </math> </span></p>

Option 3 - <p><span class="mathml" contenteditable="false"> <math> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>n</mi> <mo>+</mo> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </mrow> </mrow> </msup> </math> </span></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <msup> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mi>n</mi> <mo>/</mo> <mn>2</mn> </mrow> </mrow> </msup> </math> </span></p>

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Alok Kumar Singh | Contributor-Level 10

4 months ago

Correct Option - 3

Detailed Solution:

Given $F \propto \frac{1}{r^{n}}$

$m ω^{2} R = \frac{K}{R^{n}}$

$? \frac{m . 4 π^{2} R}{T^{2}} = \frac{K}{R^{n}}$

$? T^{2} \propto R^{n + 1}$

$T \propto R^{\frac{n + 1}{2}}$

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