Objects in elliptical orbits sweep out equal areas in equal times. This implies that the orbital speed of planet around sun is not uniform. It moves fastest at the point closest to the sun (known as the perihelion) and slowest at the point farthest away (known as aphelion). In this problem, we will calculate the difference in this speed using Pluto as an example. Pluto's orbit has an eccentricity e = 0.25. Its semi-major axis is 5.9 times 10^9 km. Determine the distance (D_aphelion) between Pluto and the sun at aphelion. You should be able to determine this using just the semi-major axis and the eccentricity. Determine the distance (D_perihelion) between Pluto and the sun at perihelion. Again, you should be able to determine this using just the semi-major axis and the eccentricity. We now want to determine the ratio between Pluto's velocity at aphelion and perihelion: v_aphelion/v_perihelion. To do this you need to find the area swept out by Pluto's orbit. This can be approximately described as a triangle with: Area = 1/2 D v_i where D is the distance from the Sun. v is velocity, and t is time. Remember that Kepler's Second Law essentially states that planets sweep out "equal areas in equal times". This means that the area swept out in some fixed time interval (Delta t) is the same at perihelion as it is at aphelion. Therefore we can say: 1/2 D_perihelion v_perihelion Delta l = 1/2 D_aphelion v_aphelion Delta l Using Equation (2), derive an expression for v_aphelion/v_perihelion. Given that Pluto's minimum orbital velocity is 3.7 km s^-1, determine values for v_aphelion, and v_perihelion.