Question

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.6 multiply 1010 m (inside the orbit of Mercury), at which point its speed is 9.3 multiply 104 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6 multiply 1012 m from the Sun? (This is the approximate distance of Pluto from the Sun.)

Answer 1

We will apply the concepts related to energy conservation to develop this problem. In this way we will consider the distances and the given speed to calculate the final speed on the path from the sun. Assuming that the values exposed when saying 'multiply' is scientific notation we have the following,

$d_1 = 4.6*10^{10}m$

$v_i = 9.3*10^4m/s \rightarrow \text{Initial velocity comet}$

$d_2 = 6*10^{12}m$

The difference of the initial and final energy will be equivalent to the work done in the system, therefore

$E_f = E_i +W$

$K_f +U_f = K_i +U_i + 0$

$\frac{1}{2} mv_f^2+\frac{-GMm}{d_2} = \frac{1}{2} mv_i^2+\frac{-GMm}{d_1}$

Here,

m = Mass

$v_f$ = Final velocity

G = Gravitational Universal Constant

M = Mass of the Sun

m = Mass of the comet

$v_i$ = Initial Velocity

Rearranging to find the final velocity,

$v_f = \sqrt{v_i^2+2GM(\frac{1}{d_2}-\frac{1}{d_1})}$

Replacing with our values we have finally,

$v_f = \sqrt{(9.3*10^4)+2(6.7*10^{-11})(1.98*10^{30})(\frac{1}{6*10^{12}}-\frac{1}{4.6*10^{10}})}$

$v_f = 75653.9m/s$

Therefore the speed is 75653m/s