Question

A space probe near the earth has values of E, L and m such that L/m = 7.8E10 m²/s and 2E/m = 0. Find the eccentricity of this orbit and the radius of the cloest point of the orbit to the earth and classify it as elliptical parabolic or hyperbolic

Answer 1

The space probe in question has a parabolic orbit with an eccentricity of 1, which arises from the given condition that its total specific energy is zero.

Step-by-step explanation:

To determine the eccentricity and the type of orbit for a space probe near Earth, you can use the given values for angular momentum per unit mass (L/m) and the specific energy (E/m). Given that 2E/m = 0, this implies that the total specific energy of the orbit is zero. Therefore, according to orbital mechanics, an orbit with zero total specific energy must be parabolic.

The eccentricity of a parabolic orbit is exactly 1. To find the radius of the closest point of the orbit to Earth, known as the perigee, we use the fact that L = mwprp2, where wp is the angular velocity at perigee and rp is the perigee radius. Since L/m is given, if we can find wp, we can solve for rp. However, the provided information is insufficient to find wp, and additional data would be needed to calculate rp.

Therefore, based on the provided information, the orbit is parabolic with an eccentricity of 1, and additional data is required to determine the radius of the closest approach.