Question

Halley's Comet revolves around the sun with time period of 76 years. The aphelion distance if perihelion is given by $8.9\times 10^{10}m$, will be (Take, mass of sun $=2\times 10^{30}kg$ and $G=6.67\times 10^{-11}N{m}^3/k{g}^2)$

• A. $4.7\times 10^{11}m$
• B. $3.23\times 10^{12}m$
• C. $5.3\times 10^{12}m$
• D. $7.63\times 10^{11}m$

Answer 1

Answer: (C)

According to Kepler's third law, time period
$T^2=\frac{4{\pi }^2}{GM}{a}^3$
where, $a$ is the semi-major axis,
$\Rightarrow a={\left[\frac{76\times 86400\times 365\times 6.67\times 2\times 10^{30}}{4\times 3.14\times 3.14}\right]}^{1/3}$
$=2.7\times 10^{12}m$
Also in case of ellipse,
$2a=\text{ perihelion }+\text{ aphelion }$
$\Rightarrow \text{ Aphelion }=2a-\text{ perihelion }$
$=2\times 2.7\times 10^{12}-8.9\times 10^{10}$
$\approx 5.3\times 10^{12}m$