Question

A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit of Mercury), at which point its speed is 9.1 104 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6 1012 m from the Sun

Answer 1

Step-by-step explanation:

From the given information:

Distance
`d_i = 4.8 * 10^(10) \ m`

Speed of the comet
`V_i = 9.1 * 10^(4) \ m/s`

At distance
`d_2 = 6 * 10^(12) \ m`

where;

mass of the sun =
`1.98 * 10^(30)`

`G = 6.67 * 10^(-11)`

To find the speed
`V_f`:

Using the formula:

`E_f = E_i + W \\ \\ where; \ \ W = 0 \ \ \text{since work done by surrounding is zero (0)}`

`E_f = E_i + 0 \\ \\ K_f + U_f = K_i + U_i \\ \\ = (1)/(2)mV_f^2 + (-GMm)/(d^2) = (1)/(2)mV_i^2+ (-GMm)/(d_i) \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [ (1)/(d_2)- (1)/(d_i)\Big ]}`

`V_f = \sqrt{(9.1 * 10^(4))^2 + 2 (6.67* 10^(-11)) *(1.98 * 10^(30) ) \Big [ (1)/(6*10^(12))- (1)/(4.8*10^(10))\Big ]}`

`\mathbf{V_f =53.125 * 10^4 \ m/s}`