Question

The minimum distance of a comet from the Sun is observed to be an eighth of the radius of the Earth’s orbit (which

is assumed to be circular), and its speed at that point (the perihelion) is four times the orbital speed vE of the Earth.

The orbits of the Earth and comet are coplanar.

a) Find the comet’s speed in terms of vE when it crosses the Earth’s orbit, and the angle at which the orbits cross.

b) Will the comet subsequently escape from the Solar System (i.e. never return)? Explain your reasoning.

Answer 1

At perihelion and aphelion, the comet's velocity is perpendicular to its position vector from the Sun. Hence, at these two special points, the comet's angular momentum (around the Sun) takes the particularly simple form, l= m r v

where, l= angular momentum, r is the radius and v is the velocity.

r= distance of the comet from the sun= R (radius of the earth)/ 8

v= speed of the comet = 4 vE

According to Kepler's second law, the comet orbits the Sun with constant angular momentum.

Hence, r₀v₀ = r₁v_1,where r₀,v_0,r_1,v₁ are the perihelion distance and speed and aphelion distanceand speed respectively

therefore,

speed of the comet= (r₁v₁)/r₀

vE=(2r₁v₁)/R

If the comet has the speed greater than the escape velocity of the comet then it can easily escape from the solar system.