How Would a Halved Sun Mass Alter Earth's Orbit?

AI Thread Summary
If the sun's mass were halved, Earth's orbit would become parabolic, indicating a transition to an unbound trajectory. The total energy of the system plays a crucial role, with zero energy resulting in a parabolic orbit, while negative energy leads to bound orbits. Calculating escape velocity using the equation E = 1/2 mv^2 - GMm/r helps determine the nature of the orbit. Approximations in mass and speed complicate precise predictions, but analytical methods can clarify the relationship between radius and velocity. Ultimately, the analysis confirms that with reduced solar mass, Earth's average speed could exceed escape velocity, resulting in a parabolic trajectory.
AdkinsJr
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Homework Statement



If the mass of the sun were 1/2 it's current mass suddenly show the orbit of the Earth would be a parabola...

The Attempt at a Solution



I'm not sure what kind of approach to apply here, if the total energy equals zero then the orbit is a parabolic trajectory, if it is less than zero the the orbit is circular or elliptical, if I plug in numbers for the mass of the earth, sun, the radius and the average speed I can find escape velocity,

E=\frac{1}{2}mv^2 -\frac{GMm}{r}

and solve for the velocity v=\sqrt\frac{GM}{r}

and show that the Earth's average speed exceeds this value, but since these numbers are going to be approximations it's difficult to tell whether it should be parabola or bound orbit...E will not be exactly zero...
 
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AdkinsJr said:
and show that the Earth's average speed exceeds this value, but since these numbers are going to be approximations it's difficult to tell whether it should be parabola or bound orbit...E will not be exactly zero...
You can do it analytically, assuming Earth has a perfectly circular orbit* you can express the radius as function of the velocity (or vice versa) and calculate the total energy afterwards.

*otherwise the whole statement is not true anyway
 
That makes sense, by radius I presume you mean the orbital equations solns. Yeah I think that will simplify things down a bit. r(\theta)=\frac{\alpha}{1+e cos(\theta)}...for a circle e=o so r_c=\frac{ml^2}{k} where l=L/m and k=GMm

r_c=\frac{m(v_cr_c)^2}{GMm} if I solve for V_c I get v_c^2=\frac{GM}{r}

If I plug this into energy equation it does actually work when I plug in 1/2 M for the mass of the sun in the potential energy term.
 
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