A comet orbits the Sun in the same plane as Earth's orbit

Question

If the Earth around the Sun follows a practically circular orbit of radius equal to one astronomical unit, determine its orbital speed in astronomical units per year.

If the Earth around the Sun follows a practically circular orbit of radius equal to one astronomical unit, determine its orbital speed in astronomical units per year. A comet orbits the Sun in the same plane as Earth's orbit. The comet's closest approach to the center of the Sun is 0.5 u.a. and it has, at that point, an orbital velocity of 4pi u.a / year. What is the orientation of the comet at the moment it cuts the Earth's orbit

Answer 1

I'll just give an outline about how I might try to answer these questions, rather than answering them outright.

Answer the first question using the circular form of the Vis-Viva equation: $v=\sqrt{\frac{\mu}{r}}$, where $\mu$ is the standard gravitational parameter ($4\pi^2 AU^3/yr^2$) and $r$ is 1 AU. If your answer isn't $2\pi$ AU per year, you probably did something wrong with the units.

For the second question, try to determine the shape of the comet's orbit using the inequality for escape velocity: $v_e \ge \sqrt{\frac{2\mu}{r}}$. Once you have the orbit shape, the rest of the problem falls out pretty easily using known equations for that orbit.

Escape velocity at 0.5 AU is $$v_e=\sqrt{\frac{2*4\pi^2}{0.5}}=\sqrt{16\pi^2}=4\pi$$ which is the exact speed given in the question. That means the orbit is parabolic. From Bogan's table pulled from Bate/Mueller/White's Fundamentals of Astrodynamics, for a parabolic orbit, $$r=\frac{2q}{1+\cos{\theta}}$$ where $r$ is the distance from the central body, $q$ is the closest distance of the orbit, and $\theta$ is the angle from periapsis or true anomaly. Solving gives us $\theta=\pi/2$. The same table gives us the angle of velocity relative to the perpendicular to the radial direction: $$\phi = \theta/2 = \pi/4$$ So the orientation of the comet's velocity is at $\pi/4$ or 45 degrees off a tangent line to a circle around the Sun with radius 1 AU as the comet passes through it.

Answer 2

The solution to such problems is a very simple application of conservation laws.

The first relevant law is conservation of angular momentum, which says that for a particular orbit $$ L = mrv_{\rm tan} = {\rm constant}\ ,$$ where $m$ is the mass of the orbiting body (assumed $\ll$ the mass $M$ of the body it orbits) and $v_{\rm tan}$ is the tangential velocity component (i.e. that component perpendicular to a line between the central mass and the orbiting object) and $r$ is the distance between the orbiting object and central mass.

The second is conservation of energy which says $$ \frac{1}{2}mv^2 - \frac{GMm}{r} = {\rm constant}\ ,$$ where $v$ is the speed in the orbit, which is formed from a tangential component (discussed above) and a radial component directed along a line towards or way from the central object.

In an elliptical orbit, an object in general has both a tangential and radial velocity component, except at aphelion and perihelion, where the radial component is zero and $v = v_{\rm tan}$.

Thus if we know the speed at perihelion, we can use conservation of angular momentum to work out $v_{\rm tan}$ at any other position in the orbit and we can use conservation of energy to calculate $v$ at any other position in the orbit.

If we are in possession of the speed and one of its components, then we know which way the velocity is directed.