Question

A 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

`v_(f)=51109.88m/s`

Step-by-step explanation:

Given data

At d₁=4.5×10¹⁰m the speed of comet is Vi=9.2×10⁴m/s

At d₂=6×10¹²m the speed will Vf

The work done by surrounding is zero and the energy of system at the initial state equal the energy at final state so we can apply:

`E_(f)=E_(i)+W\\K_(f)+U_(f)=K_(i)+U_(i)+0\\1/2mv_(f)^(2)+(-GMm)/(d_(2))=1/2mv_(i)^(2)+(-GMm)/(d_(i)) \\v_(f)=\sqrt{v_(i)^2+2GMm[1/d_(2)-1/d_(1)]}`

Substitute the given values to find final velocity Vf

So

`v_(f)=\sqrt{(9.2*10^(4)m/s )^2+2(6.7*10^(-11))(1.98*10^(30))[1/6*10^(12)-1/4.5*10^(10)]} \\v_(f)=51109.88m/s`