# A comet orbits the Sun in a highly eccentric orbit.  When the comet is at perihelion, a distance d = 4 billion meters from the Sun, the comet's speed is 15000m/s.  What is the comet's speed at aphelion, where D = 20 billion meters from the Sun?  Please see attachment for problem details.

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The comet goes around the Sun in a highly elliptic orbit so that its distance from the Sun at perihelion is 9 billion meters at it is 20 billion meters at aphelion.

The velocity of the comet at a distance r from the Sun, with mass M is given by the formula `v = sqrt(GM*(2/r - 1/a))` , where G is the gravitational constant and a is the semi major axis.

At perihelion, `r = a*(1 - e)` and at aphelion `r = a*(1 + e)` .

Using the values given:

9*10^9 = a*(1 - e) and 20*10^9 = a*(1 + e)

This gives `(1 + e)/(1- e) = 20/9`

9 + 9e = 20 - 20e

11 = 29e

e = 11/29

a = `9/(1 - 11/29)*10^9 = 14.5*10^9 `

At r = 9 billion m the velocity is 15000 m/s

`15000 = sqrt(GM*(2/(9*10^9) - 1/(14.5*10^9)))`

`GM*(2/(9*10^9) - 1/(14.5*10^9)) = 15000^2`

`G*M = 15000^2/(2/(9*10^9) - 1/(14.5*10^9))`

At r = 20*10^9

v = `sqrt(15000^2/(2/(9*10^9) - 1/(14.5*10^9))*(2/(20*10^9) - 1/(14.5*10^9)))`

= 6750

The comet has a velocity of 6750 m/s at the aphelion.

Approved by eNotes Editorial Team