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

Question

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.

Answer 1

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.

Answer 2

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:

• "Why should the trick solution to the problem of how long it takes to fall into the sun work?". (The "trick solution" is exactly that of using Kepler's Third Law.)
• "Will an object fall towards the sun if it not orbiting the sun?"
• "What if the Earth stops in its orbit and instead falls straight towards the Sun?"
• "Calculations of Earth falling into the Sun if it suddenly stopped orbiting?"

Answer 3

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.