Question

Suppose astronomers were to discover a new comet at an aphelion of 40.0 AU, with a transverse velocity of v₀= 941 m/s

What

Answer 1

The comet with mass m moves in an elliptical path around the Sun with mass M, where the semimajor axis has length a, the aphelion distance is the furthest the comet is from the Sun and perihelion is the closest the comet gets to the Sun.

Step 1) To find the angular momentum L of the comet at aphelion (angular momentum is conserved anyway so it'll be the same at any point, but we'll calculating it using the values given at aphelion), use the formula where r is the vector from the Sun to the comet and p is the momentum of the comet. The momentum of the comet is , where v is the linear speed of the comet. At aphelion, the direction of the comet is perpendicular to the distance from the Sun to the comet, so the angular momentum reduces to .

Step 2) We're interested in the angular momentum per unit mass, so instead we want to calculate . Since , here . Plug and into .

The angular momentum per unit mass is .

Step 3) The total mechanical energy is constant anywhere around the orbit, and is equal to the sum of the potential and kinetic energies of the comet. If the comet is a distance r away from the Sun, the potential energy U is where G is the gravitational constant. The kinetic energy is , so the total energy is: . The total energy per unit mass then is .

Step 4) Again since the total mechanical energy is constant, we can choose to use the values given at aphelion to calculate the total energy. Plug in , , and into .

The total energy per unit mass is . It is negative which means that comet is gravitationally bound: it doesn't have enough kinetic energy to overcome the gravitational pull from the Sun.

Step 5) Another formula for the total mechanical energy for an object in an elliptical orbit is (recall a is the semimajor axis of the ellipse). The aphelion radius is related to the eccentricity e by the equation . Rearrange this to get . Plug this into to get .

Step 6) Rearrange the above equation for the total energy to solve for the eccentricity e.

Step 7) Plug in , , , .

Elliptical orbits have eccentricities between 0 and 1, so this value makes sense.

Step 8) Since and , solve for a to get and plug this into to get .

Now plug in and and solve for .

The closest approach (perihelion) is at , which is less than aphelion as it should be.

Step 9) To find the comets speed at perihelion, recall that total angular momentum is conserved. In steps 1 and 2 we said that angular moment is constant and the angular momentum per unit mass could be found using the equation . Plug in and and solve for the speed v at perihelion.

The speed at perihelion is . Notice this is much faster than at aphelion, which is as it should be: the further from the Sun, the slower it goes, and the closer to the Sun, the faster it goes. This is because angular momentum is conserved.