Get Answers to all your Questions

header-bg qa

An object is allowed to fall from a height R above the earth, where R  is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be

Option: 1

\sqrt{2 g R}


Option: 2

\sqrt{\frac{g R}{2}}


Option: 3

2 \sqrt{g R}


Option: 4

\sqrt{g R}


Answers (1)

best_answer

Use work energy theorem

\Delta \mathrm{KE}=\mathrm{w}_{\mathrm{g}}

\frac{1}{2} m v^{2}-0=-\left[\mathrm{u_f} -\mathrm{u}_{\mathrm{i}}\right]
\frac{1}{2} m v^{2}=-\left[-\frac{G M m}{\mathrm{R}}-\left(-\frac{\mathrm{GMm}}{2 \mathrm{R}}\right)\right]
\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{2 \mathrm{R}}
            =\frac{\mathrm{GMm}}{\mathrm{R}}\left(\frac{2-1}{2}\right)

\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{GMm}}{2 \mathrm{R}}
\mathrm{V}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}
\mathrm{V}=\sqrt{g R}\left(\mathrm{GM}=\mathrm{gR}^{2}\right)

Posted by

Riya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE