Question

Given the perihelion distance, p, and aphelion distance, q, for an elliptical orbit, show that the velocity at perihelion, vp, is given by vp=√2GMSun(q+p)qp. (Hint: Use conservation of angular momentum to relate vp and vq, and then substitute into the conservation of energy equation.)

(a) ( vₚ = sqrt{frac{2GMSun(q+p)}{qp}} )
(b) ( vₚ = sqrt{frac{GMSun(p+q)}{2pq}} )
(c) ( vₚ = sqrt{frac{2GMSun(p+q)}{pq}} )
(d) ( vₚ = sqrt{frac{GMSun(q+p)}{2pq}} )

Answer 1

Using conservation of angular momentum and energy, we can derive the expression for the velocity at perihelion to be √(2G x MSun x (q + p) / (q x p)).

Step-by-step explanation:

To find the velocity at perihelion, vp, given the perihelion distance, p, and aphelion distance, q, we use the conservation of angular momentum and the conservation of energy. By conservation of angular momentum, Lp = Lq, and because L = r x m x v (where r is the radius, m is the mass, and v is velocity), we have m x p x vp = m x q x vq. By conservation of energy, K.E. + P.E. = constant, we have ½ m x vp² - G x MSun x m / p = ½ m x vq² - G x MSun x m / q. After substituting vq from the angular momentum equation into the energy equation and solving for vp, we obtain the expression vp = √(2G x MSun x (q + p) / (q x p)).

Therefore, out of the given options, the correct expression for the velocity at perihelion is (c) vp = √(2G x MSun x (p + q) / (p x q)).