Question

The minimum distance of a satellite from the centre of the earth is 3R, where R is the radius of 2GM What is the earth. The mass of earth is M. The maximum speed of the satellite is $\sqrt{\frac{2GM}{3R}}$ radius of curvature at the point of minimum distance?

• A. 6 R
• B. 4 R
• C. 3 R
• D. $\frac{9R}{2}$

Answer 1

The correct option is A 6 R

$\begin{array}{c}F=\frac{G\left(M\right)m}{{\left(3R\right)}^2}\\ \Rightarrow a=\frac{GM}{9R^2}\\ \therefore \frac{V^2}{R^{\prime }}=a\\ \Rightarrow \frac{2GM}{3R.R^{\prime }}=\frac{GM}{9R^2}\\ \Rightarrow R^{\prime }=6R\\ Hence,\\ optionAiscorrectanswer.\end{array}$