Halley’s comet has a period of 76, had distance of closest approach to the sun equal to $8.9 \times {10^{10}}{\rm{m}}$. the comet’s farthest distance from the sun if the mass of sun is $2 \times {10^{30}}kg\,and\,G = 6.67 \times {10^{11}}$ in MKS units is

JEE PHYSICS MCQ: GRAVITATION

1 Medium

1 of 1

Application +4 | -1

Halley’s comet has a period of 76, had distance of closest approach to the sun equal to

8.9 \times {10^{10}}\mathrm{m}

. the comet’s farthest distance from the sun if the mass of sun is

2 \times {10^{30}}kg\,and\,G = 6.67 \times {10^{11}}

in MKS units is

View answer and explanation

Correct option: C

5.3 \times {10^{12}}\mathrm{m}

Option C (

5.3 \times {10^{12}}\mathrm{m}

) is correct because it is calculated using Kepler's third law and the properties of an ellipse.

Practice All Chapter MCQs →

11.2\mathrm{\;\;km}{\mathrm{s}^{ - 1}}

{45^0}

v_e

\theta