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 $\left.G=6.67 \times 10^{-11} Nm ^3 / kg ^2\right)$

• 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
$T^2=\frac{4 \pi^2}{G M} a^3$
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,
$ 2 a=\text { perihelion }+\text { aphelion }$
$\Rightarrow \text { Aphelion }=2 a-\text { perihelion }$
$ =2 \times 2.7 \times 10^{12}-8.9 \times 10^{10}$
$ \approx 5.3 \times 10^{12} m $