Fried EarthOur earth has orbital motion, revolving once around the sun in about 365 days. Suppose that this orbital motion suddenly stopped completely. How long would it take for the earth to plunge along a straight line into the sun?
This prolem is an example of the advantages, and practical disadvantages of using insightful methods for homework problems. It was an assigned homework problem in classical mechanics when I was a university student in 1957. Always looking for ways to avoid excess work, I deliberately tried to find the answer by simple means. However, my professor did not take kindly to the solution I submitted, though I was quite proud of it. He wanted it done in the conventional "textbook" manner, just to show that I had read the textbook and mastered its methods. So I got no points for my ingenuity and insight.
Many years later I happened to meet the author of the textbook, so I related this anecdote to him. He said, "That's an old problem that has been in many textbooks. You did it the way I hoped students would discover. Why use a sledgehammer to crack a walnut?"
With that warning, look for a simple way to arrive at the answer. Use calculus if you like, but check your answer with a simpler non-calculus method. One of my profs used to say "Most problems in undergraduate physics that are done using calculus could also be done without explicit use of calculus—if you are clever enough."
SolutionKepler's law applies to planetary orbits, whether they be of circular, or elliptical shape. It says that T22/T12 = R23/R13, where T is the period of an orbit and R is its semi-major axis. The semi-major axis is the average of the planet's maximum and minimum distances from the sun. For the earth orbit, the sun is at one focus of the ellipse.
Let the earth's mean radius be R1. Now, if the earth's orbital momentum were suddenly reduced (without exerting anything but a tangential stopping force on the earth), it would fall straight to the sun. This straight fall can be considered 1/2 of a degenerate elliptical orbit with major axis equal to R1. Its semi-major axis is R1/2 (the average of R1 and zero). Its period will be designated T2.
So: T22/T12 = (R1/2)3/R13 = (1/2)3
And therefore, T2 = T1/23/2 = 0.353 year, and the time to fall into the sun is 1/2 of that, or 0.176 years or 64.52 days—a bit over two months.
So what fault had my first professor found with this elegant solution? Several. I had not established and justified that a straight line trajectory into the sun is really a degenerate ellipse. To do that I probably should have taken the limit as an ellipse's minor axis decreased to zero. Had I dealt with the fact that my crash orbit passes through the center of the sun, where its distance from the sun's center is zero, and the gravitational force there is infinite? Well actually it isn't infinite there since the sun isn't a point object, and the earth would have vaporized long before reaching the center. Nonetheless, if such a catastrophe happened, we can predict that the earth would take a bit over two months to make its plunge to a fiery death, giving us some time to enjoy the show before frying to a crisp.