Question

comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.5 × 1010 m (inside the orbit of Mercury), at which point its speed is 9.2 × 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? (This is the approximate distance of Pluto from the Sun.)

Answer 1

Given Data:

At distance d₁ = 4.5×10¹⁰ m, speed of comet is
`v_(i)` = 9.2×10⁴ m/s.

At distance d₂ = 6×10¹² m, speed of comet will be
`v_(f)` = ?

As there is no work done from the outside, so the initial energy of the system will be equal to the final state of energy.

`E_(f) = E_(i)+W\\ K_(f) +U_(f) =K_(i) +U_(i) +0`

`(1)/(2)mv^(2) _(f) + (-GMm)/(d_(2) ) = (1)/(2)mv^(2) _(i) + (-GMm)/(d_(1) )`

`v_(f) = \sqrt{v^(2) _(i)+2GM[(1)/(d_(2)) -(1)/(d_(1) ) ] }`

Now by putting the values in the above equation we can find out
`v_(f)`. Here M is the mass of sun.

`v_(f) = \sqrt{(9.2*10^(4))^(2) + 2(6.7*10^(-11))(1.98*10^(30))[(1)/(6*10^(12))-(1)/(4.5*10^(10) ) } } ]`

= 51109.88 m/s