- Home
- Reference
- Education
- Directories
- Science eStore
- Job Board
- Ask The Experts
- eGreetings
- Fun
- About Us

Welcome to PhysLink.com - Your physics and astronomy online portal. Stay a while! Check out our extensive library of educational and reference materials. Also, check out our fun section!

QuestionLets just say a huge asteroid hits the earth at the right speed and with the right amount of force and stops it's orbit around the sun. How long would it take for the earth to plummet into the sun? Asked by: Mike Assad AnswerThere are two ways to go about this question... One is, bite the bullet and solve the differential equation, plug the values in, and get the answer. The other is, use known quantities and some knowledge about orbits. The equation of motion for a body which has no angular momentum about the sun, as described, is: r'' = -GM/r ^{2}
[G is gravitational constant, M is mass of sun, r is distance of body from the sun] Where the primes denote differentiation with respect to time. One would integrate this twice, the first step gives: r' = - sqrt (2GM) * sqrt(1/r - 1/R) Where R is the initial distance from the sun, here it would be about the distance of the earth from the sun. Second integration gets a little trickier, yet still doable; and it yields the following equation: r*R*sqrt(1/r - 1/R) + R ^{3/2}arctan(sqrt(R/r -1)) = sqrt(2GM)*t
Where t is the time measured from the time the body started its motion at distance R. Now, we want to find t when r goes to zero. It turns out easier than it looks. The first term on the left vanishes, while the arctangent's argument becomes infinite, and the arctangent of infinity is Pi/2, we get, after some massaging: t = sqrt( ^{2}*R^{3}/(8*G*M))
Is the time it will take to fall into the sun, ignoring the motion of the sun, and the size of the sun... Now how do we evaluate this? The earth's period around the sun is given by: T = sqrt(4* ^{2}*R^3/(G*M))
And of course, it is a year. Comparing the two, we see that... t = T/(4*sqrt(2)) Which is roughly 64 days and 12 hours,
or just over two month's time to say goodbye to
the cruel world, however, it might get considerably hot
towards the end...
Now, the other, less math-more-physics method. We know the period of any orbit is dependent only on the semimajor axis; as long as it is an ellipse with the sun at one focus. We can take this problem to be one where the orbit is an extremely flat ellipse. This puts the focus at one end, while the body starts the motion at the other end. So, what we need to calculate here is the HALF of the period of an orbit for which the semimajor axis is HALF of R, if you follow the argument. So, the expression is: t = (1/2)*sqrt(4* ^{2}*(R/2)^{3}/(GM))
which reduces to: t = ( ^{2}*R^{3}/(8*G*M))
The same expression as before. Also note that the mass of the asteroid does not enter into the equation, as long as it is not so huge to make the mass of the earth after the collision comparable to the mass of the sun, in which case we'll have to consider the movement of the sun as well in the derivation. (This can be taken care of just by using the reduced mass, but that's another story.) Answered by: Yasar Safkan, Ph.D. M.I.T., Software Engineer, Istanbul, Turkey Your proposed asteroid would need to have the same momentum as the mass of the Earth multiplied by its orbital velocity, but in the opposite direction. One example would be another Earth-sized object travelling at 18 miles/second. A lower mass with higher velocity would also do, but you can easily imagine any such collision completely destroying both objects with little left of the Earth to fall into the Sun. Answered by: Paul Walorski, B.A., Part-time Physics Instructor |

Loading

(