<p>I'll just give an outline about how I might try to answer these questions, rather than answering them outright.</p>
<p>Answer the first question using the circular form of the <a href="https://en.wikipedia.org/wiki/Vis-viva_equation" rel="nofollow noreferrer">Vis-Viva</a> equation: <span class="math-container">$v=\sqrt{\frac{\mu}{r}}$</span>, where <span class="math-container">$\mu$</span> is the <a href="https://en.wikipedia.org/wiki/Standard_gravitational_parameter" rel="nofollow noreferrer">standard gravitational parameter</a> (<span class="math-container">$4\pi^2 AU^3/yr^2$</span>) and <span class="math-container">$r$</span> is 1 AU.  If your answer isn't <span class="math-container">$2\pi$</span> AU per year, you probably did something wrong with the units.</p>
<p>For the second question, try to determine the shape of the comet's orbit using the inequality for escape velocity: <span class="math-container">$v_e \ge \sqrt{\frac{2\mu}{r}}$</span>.  Once you have the orbit shape, the rest of the problem falls out pretty easily using known equations for that orbit.</p>
<blockquote class="spoiler">
<p> Escape velocity at 0.5 AU is
 <span class="math-container">$$v_e=\sqrt{\frac{2*4\pi^2}{0.5}}=\sqrt{16\pi^2}=4\pi$$</span>

 which is the exact speed given in the question. That means the orbit
 is parabolic.  From  <a href="http://www.bogan.ca/orbits/kepler/orbteqtn.html" rel="nofollow noreferrer">Bogan's table</a> pulled from
 Bate/Mueller/White's Fundamentals of Astrodynamics, for a parabolic
 orbit, <span class="math-container">$$r=\frac{2q}{1+\cos{\theta}}$$</span> where <span class="math-container">$r$</span> is the distance from
 the central body, <span class="math-container">$q$</span> is the closest distance of the orbit, and
 <span class="math-container">$\theta$</span> is the angle from periapsis or true anomaly. Solving gives us
 <span class="math-container">$\theta=\pi/2$</span>.

 The same table gives us the angle of velocity relative to the
 perpendicular to the radial direction: <span class="math-container">$$\phi = \theta/2 = \pi/4$$</span>

 So the orientation of the comet's velocity is at
 <span class="math-container">$\pi/4$</span> or 45 degrees off a tangent line to a circle around the Sun
 with radius 1 AU as the comet passes through it.</p>
</blockquote>

CC BY-SA 4.0
2021-03-06T19:10:59.623Z
The answer is very simple (5 lines) and does not need the vis-viva equation or the equation of an ellipse.
ProfRob
CC BY-SA 4.0
2021-03-05T19:52:52.747Z
2021-03-09T21:48:36.260Z
2021-03-09T21:48:36.260Z
Connor Garcia
<p>The solution to such problems is a very simple application of conservation laws.</p>
<p>The first relevant law is conservation of angular momentum, which says that for a particular orbit
<span class="math-container">$$ L = mrv_{\rm tan} = {\rm constant}\ ,$$</span>
where <span class="math-container">$m$</span> is the mass of the orbiting body (assumed <span class="math-container">$\ll$</span> the mass <span class="math-container">$M$</span> of the body it orbits) and <span class="math-container">$v_{\rm tan}$</span> is the tangential velocity component (i.e. that component perpendicular to a line between the central mass and the orbiting object) and <span class="math-container">$r$</span> is the distance between the orbiting object and central mass.</p>
<p>The second is conservation of energy which says
<span class="math-container">$$ \frac{1}{2}mv^2 - \frac{GMm}{r} = {\rm constant}\ ,$$</span>
where <span class="math-container">$v$</span> is the <em>speed</em> 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.</p>
<p>In an elliptical orbit, an object in general has <em>both</em> a tangential and radial velocity component, <em>except</em> at aphelion and perihelion, where the radial component is zero and <span class="math-container">$v = v_{\rm tan}$</span>.</p>
<p>Thus if we know the speed at perihelion, we can use conservation of angular momentum to work out <span class="math-container">$v_{\rm tan}$</span> at any other position in the orbit and we can use conservation of energy to calculate <span class="math-container">$v$</span> at any other position in the orbit.</p>
<p>If we are in possession of the speed and one of its components, then we know which way the velocity is directed.</p>

CC BY-SA 4.0
2021-03-07T11:00:04.987Z
2021-03-07T11:23:25.550Z
2021-03-07T11:23:25.550Z
ProfRob
<p>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.</p>
<p>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</p>

CC BY-SA 4.0
2021-03-05T18:45:51.650Z
Sounds like a homework/test question.
slowerthanstopped
CC BY-SA 4.0
2021-03-05T23:29:00.547Z
@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
CC BY-SA 4.0
2021-03-05T23:30:53.243Z
@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
CC BY-SA 4.0
2021-03-06T00:06:56.407Z
@AlbertWesker see for example the helpful comments under [this closed question](https://astronomy.stackexchange.com/q/41788/7982)
uhoh
CC BY-SA 4.0
2021-03-06T00:12:02.410Z
uhoh what slowerthanstopped meant is that we’re not here to answer homework/test questions…
Pierre Paquette
CC BY-SA 4.0
2021-03-06T19:14:13.620Z
Apply conservation laws.
ProfRob
<gravity><comets><positional-astronomy>
A comet orbits the Sun in the same plane as Earth's orbit
Albert Wesker
# A comet orbits the Sun in the same plane as Earth's orbit

(Asked by: username_0 on 2021-03-05)

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

**Tags:** gravity, comets, positional-astronomy

> **Comments:**
> **username_1**: Sounds like a homework/test question.  
> **username_2**: @username_1 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.  
> **username_2**: @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.  
> **username_2**: @AlbertWesker see for example the helpful comments under [this closed question](https://astronomy.stackexchange.com/q/41788/7982)  
> **username_3**: username_2 what username_1 meant is that we’re not here to answer homework/test questions…  
> **username_4**: Apply conservation laws.  

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## 2 answers

### Answer by username_5 on 2021-03-05. Score: 3

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](https://en.wikipedia.org/wiki/Vis-viva_equation) equation: $v=\sqrt{\frac{\mu}{r}}$, where $\mu$ is the [standard gravitational parameter](https://en.wikipedia.org/wiki/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](http://www.bogan.ca/orbits/kepler/orbteqtn.html) 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.

> **Comments:**
> **username_4**: The answer is very simple (5 lines) and does not need the vis-viva equation or the equation of an ellipse.  

---

### Answer by username_4 on 2021-03-07. Score: 1

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.


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