The minimum and maximum distance of a satellite from the centre of the earth are 2R and 4R, respectively, where R is the radius of earth and M is the mass of earth. The radius of curvature at the point of maximum distance is

Option 1 - <math> <mrow> <mfrac> <mrow> <mn>8</mn> <mi>R</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

Option 2 - <math> <mrow> <mfrac> <mrow> <mn>4</mn> <mi>R</mi> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

Option 3 - <math> <mrow> <mfrac> <mrow> <mn>3</mn> <mi>R</mi> </mrow> <mrow> <mn>8</mn> </mrow> </mfrac> </mrow> </math>

Option 4 - <math> <mrow> <mfrac> <mrow> <mn>3</mn> <mi>R</mi> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> </mrow> </math>

11 Views|Posted 5 months ago

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Answered by

Alok Kumar Singh | Contributor-Level 10

5 months ago

Correct Option - 1

Detailed Solution:

$\frac{1}{2} m V_{a}^{2} - \frac{G M m}{2 x}$

= $\frac{1}{2} m \frac{V_{a}^{2}}{4} - \frac{G M m}{4 x}, V_{a} = \sqrt[]{\frac{2}{3} \frac{G M}{x}}$

ROC at point – B

$R O C = \frac{V_{b}^{2}}{a ⊥} = \frac{\frac{2 G M}{3 R} * \frac{1}{4}}{\frac{G M}{{(4 R)}^{2}}} = \frac{8 R}{3}$

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