JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2020
- Hard
- Numerical
An asteroid is moving directly towards the centre of the earth. When at a distance of \(10 \mathrm{R}\) (\(R\) is the radius of the earth) from the earths centre, it has a speed of \(12 \;\mathrm{km} / \mathrm{s}\). Neglecting the effect of earths atmosphere, what will be the speed of the asteroid when it hits the surface of the earth (escape velocity from the earth is \(11.2 \;\mathrm{km} / \mathrm{s}\) ) \(?\) Give your answer to the nearest integer in \(km/s\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(16\)
Step-by-step Solution
Detailed explanation
\(\mathrm{U}_{1}+\mathrm{K}_{1}=\mathrm{U}_{2}+\mathrm{K}_{2}\)
\(-\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{10 \mathrm{R}}+\frac{1}{2} \mathrm{mv}_{0}^{2}=-\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}^{2}\)
\(+\frac{9}{10} \times \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}}+\frac{1}{2} \mathrm{mv}_{0}^{2}=\frac{1}{2} \mathrm{mv}^{2}\)
\(\frac{9}{10} \times \frac{1}{2} \mathrm{M} \times \mathrm{v}_{\mathrm{e}}^{2}+\frac{1}{2} \mathrm{mv}_{0}^{2}=\frac{1}{2} \mathrm{mv}^{2}\)
\(\mathrm{v}^{2}=\frac{9}{10} \mathrm{v}_{e}^{2}+\mathrm{v}_{0}^{2}\)
\(=\frac{9}{10} \times(11.2)^{2}+(12)^{2}\)
\(\mathrm{v}^{2}=112.896+144\)
\(\mathrm{v}=16.027\)
\(\mathrm{v}=16 \mathrm{km} / \mathrm{s}\)
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