# Which type of trajectory would an object take in the below stated case?

What type of trajectory would an object take if it is suddenly stopped by some external force while revolving in a circular orbit around the sun?(Neglect the radius of the sun i.e radius of the circular orbit is much greater than the radius of the sun).My teacher said it would take an thin elliptical path with semi-major axis r/2.But why would it take an elliptical path.Can anyone explain how this happens.

• Probably you understood him not. From rest relative to the sun it would fall to the surface of the sun in a straight line. The time would be the same as for half the circle. Commented Aug 3 at 14:30
• This is an extremely poor question. Why do you think the trajectory will be different from one starting from rest? Commented Aug 4 at 10:25
• I think you didn't understand the question.The object in the above case sure takes a elliptical path.I am asking the reason for it.Refer sources if you want to confirm.Please don't downvote unnecessarily Commented Aug 7 at 14:15

If the sun is the only mass attracting the object, and it isn't moving in your reference frame, and then you release the small object "from rest" the object will follow a linear trajectory straight into the sun's center. In order to have an elliptical orbit, the small object needs to have some non-zero velocity in a direction other than the unit vector between the two bodies.

• If "very far" is large enough the Sun will have moved by the time the object reaches it. I don't know whether it can move far enough for the object to miss it though. Commented Aug 3 at 14:57
• I modified the question.Please refer the modified one and answer accordingly. Commented Aug 4 at 7:19
• It's still pretty much the same question and, as such, it has the same answer. Commented Aug 4 at 16:20

My guess is that your teacher was probably trying to explain that taking the straight line path as the limit of an infinitely narrow ellipse makes it much simpler to calculate the time it takes for the object to fall into the Sun, since using Kepler's Third Law is much easier that direct integration of the gravitational acceleration along the straight line path.

For more details, look at "Earth Falling Towards the Sun" and compare the length of the "Overall solution" (that does the integration) to the length of the "Overall solution (Upper-secondary level)" (that uses Kepler's Third Law). The direct solution involves this unpleasant integral: $$\int{\sqrt{\frac{r}{R_{\mathrm{ES}}-r}}}\mathrm{d}r=\frac {\sqrt{\frac{r}{R_{\mathrm{ES}}-r}} \left(\sqrt{r} \left(r-R_{\mathrm{ES}} \right) + R_{\mathrm{ES}}\sqrt{R_{\mathrm{ES}}-r} \mathrm{arctg}\,{\sqrt{\frac{r}{R_{\mathrm{ES}}-r}}} \right) }{\sqrt {r}}+c,$$ In comparison, using the limit of an ellipse with Kepler's Third Law immediately tells us that the fall time is simply a quarter of the orbital period: $$T = 2\pi\sqrt{\frac{a^3}{GM}}$$

where $$a$$ is the orbit's semi-major axis, $$G$$ is the gravitational constant, and $$M$$ is the reduced mass.

This is a classic problem discussed in many places, including these Physics Stack Exchange questions:

• Thanks for the solution.I understood it clearly now Commented Aug 7 at 14:17

The trajectory of a body in a central gravitational field can only be a conic section in a plane containing the center of the force. The latter occupies a focus of the conic. Position and velocity at a given time completely fix the trajectory.

Therefore if a body is suddenly stopped by a force at a certain position with respect a center of gravity, the motion, along a suitable conic, is from then on determined by the position and the fact that there the velocity vanishes. It does not matter what was the motion before the stop, if assuming that after the stop the only force acting on the body is the gravitational one.

So we have to look for a solution of the equations of motion such that, at a certain time, the point is in a certain position (different of the center) and with zero velocity. Along every permitted trajectory, in the considered case, the velocity of the point vanishes nowhere (because it defines the tangent to the conic), barring the limit case of an ellipse with a degenerate axis. In this case the motion runs along a (finite) segment passing through the center of force, and the zero velocity is attained at the extreme points of the segment where the direction of the motion is reversed.

In the concrete realistic case, the body will fall along a straight line towards the center of force, the sun, and presumably it will stop there.

• I modified the question.Please refer the modified one and answer accordingly. Commented Aug 4 at 7:20
• You should say what you modified. For me all the answers apply to yor question as vI read it now. Commented Aug 4 at 14:14
• It applies to the new version too: the trajectory is a straight line from the initial position to the sun. Commented Aug 4 at 14:18