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

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

• Sounds like a homework/test question. Mar 5 at 18:45
• @slowerthanstopped your comment is not helpful to the OP because they don't know the significance of "sounds like a..." and so don't know what action to take. Comments should be directed to the post author and be helpful to them and actionable; they should contain an explanation of what you think should be done to improve the post. Stack Exchange does not require people to know anything about the site before they start posting, its up to us to help new users learn about the site.
– uhoh
Mar 5 at 23:29
• @AlbertWesker Welcome to Stack Exchange! Homework questions are not off-topic here, but they need to include some discussion of what you have tried so far, and where you are stuck. The more information you can add to the question the better and more helpful the answers can be.
– uhoh
Mar 5 at 23:30
• uhoh what slowerthanstopped meant is that we’re not here to answer homework/test questions… Mar 6 at 0:12
• Apply conservation laws. Mar 6 at 19:14

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.

• The answer is very simple (5 lines) and does not need the vis-viva equation or the equation of an ellipse. Mar 6 at 19:10

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.