Question

For a given earth satellite with given angular momentum l, show that the distance of closest approach rₘᵢₙ on a parabolic orbit is half the radius of the circular orbit.

Answer 1

To show that the distance of the closest approach rₘᵢₙ on a parabolic orbit is half the radius of the circular orbit, we need to consider the angular momentum of the satellite. By comparing the equations for angular momentum in circular and parabolic orbits, it can be shown that the distance of the closest approach is half the radius of the circular orbit. This relationship holds for satellites in elliptical orbits as well.

Step-by-step explanation:

To show that the distance of the closest approach rₘᵢₙ on a parabolic orbit is half the radius of the circular orbit, we need to consider the angular momentum of the satellite.

The angular momentum l is given by l = mvr, where m is the mass of the satellite, v is its velocity, and r is the distance from the satellite to the center of the Earth.

For a circular orbit, the velocity can be expressed as v = √(GM/r), where G is the gravitational constant and M is the mass of the Earth. Substituting this into the angular momentum equation gives l = m√(GMr).

For a parabolic orbit, the eccentricity e is equal to 1. The distance of the closest approach rₘᵢₙ can be found by rearranging the equation for the angular momentum to solve for r: r = l²/(G^2M^2m²).

Substituting e = 1, we get rₘᵢₙ = l²/(2GMm²).

Comparing rₘᵢₙ to the radius of the circular orbit r, we can see that rₘᵢₙ = r/2, as required.