Answer to Question #318462 in Physics for Jiren

Question

Haley’s Comet can only be viewed once in every 76 years. How fast is it moving if it is

1.56 x 1020 m away from the Earth? If its mass is about 3,000 kg, how much force is keeping it in orbit?

Answer 1

The semi-axis of Haley’s Comet orbit is given by

$$a=\sqrt[3]{\frac{GMT^2}{4\pi^2}}$$

$$a=\sqrt[3]{T^2}=\sqrt[3]{(76\: \rm yr)^2}=17.9\:\rm AU$$

The conservation of energy gives

$$v=\sqrt{GM\left(\frac{2}{r}-\frac{1}{a}\right)}$$

$$v=\sqrt{1.33*10^{20}\left(\frac{2}{1.56*10^{12}}-\frac{1}{17.9*1.5*10^{11}}\right)}$$

$$v=1.1*10^4\:\rm m/s$$

The force

$$F=\frac{GMm}{r^2}=\frac{1.33*10^{20}*3000}{(1.56*10^{12})^2}=0.16\:\rm N$$