Hippocampus Physics: Planetary Motion

[Print]

Now let us look at the motion of a satellite in an elliptical orbit around the Earth, and try to figure out its velocity at any point on its path. In this case, the gravitational force acting on the satellite is a central force, always acting along the radial vector towards the Earth. The torque acting on the satellite due to this force, F is equal to the cross product of r and F, which is equal to zero since r and F are parallel to each other. But remember that torque is equal to the derivative of angular momentum with respect to time. If the time rate of change of a quantity is equal to zero, that quantity is constant. So the angular momentum L of the satellite is always constant. Since the angular momentum L is equal to the cross product of the radial vector and the momentum vector, a constant angular momentum means that the motion of the satellite is restricted to the plane formed by r and p, or equivalently by r and velocity v. At the apogee and perigee of the elliptical orbit, the velocity vector v is perpendicular to the radial vector. Therefore, the sine of the angle is equal to 1, and at those locations, the angular momentum is equal to the product of the velocity, radial distance, and mass of the satellite. Setting the values of angular momentum at these two points equal to each other, we obtain that the product of the velocity and the radial vector is the same at the perigee and the apogee. This implies that when the satellite is closest to the Earth, at the perigee, its velocity is at its maximum, since r is at a minimum there. At the apogee, since the distance is at its maximum, the velocity is at a minimum, and thus the satellite moves fastest at the perigee and slowest at the apogee. Simplifying, we obtain the value for the velocity at the perigee in terms of the velocity at the apogee. So, by setting the angular momentum of the object at a point equal to the angular momentum at the perigee or apogee and solving for the velocity can obtain the velocity of the satellite for any point on the orbit. In much the same way, the previous relations can be applied to a planet moving along an elliptical orbit around the Sun, so its velocity at any point of the orbit can be calculated in terms of its velocity at aphelion or perihelion. Let us do an exercise. The distance between Mercury and the Sun at Mercury's aphelion is 7 times 10 to the 10th meters, with a speed of 3.88 times 10 to the 4th meters per second. If Mercury's perihelion is 4.6 times ten to the 10th meters, what is the speed of Mercury at the perihelion? ...