• If the earth suddenly stopped in its orbit assumed to be circular, find the time
    that would elapse before it falls into the sun.

  • @Rushabh-Jain

    Kepler's law applies to planetary orbits, whether they be of circular, or elliptical shape. It says that T22{T_{2}}^2T22/T12{T_{1}}^2T12= R23{R_{2}}^3R23/R13{R_{1}}^3R13, 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.

    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{T_{2}}^2T22/T12{T_{1}}^2T12 = R13/23{{R_1}^{3}/2}^3R13/23/R13R_{1}^{3}R13= 1/23{1/2}^31/23

    And therefore, T2T_{2}T2= T1T_{1}T1/23/22^{3/2}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.

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