Gravitational motion problem, degrees along orbit

[DaviBones]

Homework Statement

A comet passes the sun at it's perihelion 5*10¹⁰ meters going 1*10⁵ meters/second. Take the mass of the sun to be 2*10³⁰.

a.) Will the comet return to this point? If so, how long until that happens?
b.) How long will it take to travel from perihelion to 60 degrees further along it's orbit?

Homework Equations

F_{g}=\frac{-GMm}{r^{2}}
KE=\frac{1}{2}mv^{2}
PE=\frac{-GMm}{r}
\omega=\frac{v}{r}

The Attempt at a Solution

Part a was easy. I summed the KE and PE and it was positive. I concluded that the velocity was great enough to escape orbit.

Part b, I have no clue how to approach. I've nothing in my arsenal that deals with degrees along an orbit besides perhaps conservation of angular momentum but I have been thinking about this for a couple hours now and I'm totally stumped on where to even begin. Any guidance would be greatly appreciated.

 

[vivekrai]

You didn't get any orbit by solving a, that's why you could not do (b). I think you should re think for part (a).

 

[DaviBones]

Does positive total energy not mean a closed orbit? I was taught that E<0 is a closed orbit, E=0 is a parabola, and E>0 is a hyperbola

 

[vivekrai]

Okay. Thanks for letting me know. Then probably you can measure 60 degrees arc of length with some maths of hyberbola at hand.

 

[gneill]

You could make use of the Kepler law which states that equal areas are swept out in equal times. The fact that you've got the perihelion radius and velocity gives you the specific angular momentum, h. Then

\int_0^t h\;dt = \int_0^{\nu} r^2 \; d\nu

r is given by the usual conic section formula for $r(\nu)$. Time t=0 corresponds to the instant of perihelion passage when $\nu = 0$.

You might want to look up references for the "Time of Flight Problem", as there are some nice reductions of the integral for the various orbit cases (ellipse, parabola, hyperbola).

 

[DaviBones]

Sorry, I'm having a hard time figuring out your post... Is everything in that equation what I expect it is? In other words, is "t" time, "v" velocity, and "r" the radius? If so, doesn't the left side become simply "ht", since angular momentum in this case is constant?

Also I couldn't find the conic section formula you're talking about anywhere, and looking up time of flight mostly turned up kinematics time of flight problems, which I doubt is what you're talking about, right? That's only parabolic motion as far as I could find.

 

[gneill]

Sorry, I'm having a hard time figuring out your post... Is everything in that equation what I expect it is? In other words, is "t" time, "v" velocity, and "r" the radius?

$\nu$ is the true anomaly, that is, the angle measured at the focus (Sun) from perihelion to the current location of the orbiting object.

If so, doesn't the left side become simply "ht", since angular momentum in this case is constant?

Yup!

Also I couldn't find the conic section formula you're talking about anywhere, and looking up time of flight mostly turned up kinematics time of flight problems, which I doubt is what you're talking about, right? That's only parabolic motion as far as I could find.

r = \frac{p}{1 + e\;cos(\nu)}
where p is the semi latus rectum.

You might want to locate a book on astrodynamics that discusses The Kepler Problem and determination of time of flight between locations along an orbit.