If orbital radius of earth is half then number of days in one year are
Answers
Hi rakhi I think the answer is 129 days But calculate it and prove it to yourself! You get 129 days, 2 hours, 45 minutes and 36 seconds. Use Kepler's 3rd to solve for orbital period. The equation is: T = SQRT { [4 * π^2 * r^3 ] / [GM] } Where T = Period π = pi r = Orbital distance G = Universal Gravitational Constant M = Mass of the Sun So we'll take 1 AU and split it in half for r, and look up M. I have G memorized... T = ? π = 3.14159 r = 74,798,935,500 m G = 6.67428E-11 m^3/kg-s^2 M = 1.9891E+30 kg Solve: T = SQRT { [4 * π^2 * r^3 ] / [GM] } T = SQRT { [4 * (3.14159)^2 * (74,798,935,500 m)^3 ] / [ (6.67428E-11 m^3/kg-s^2) * (1.9891E+30 kg) ] } T = SQRT { [4 * (9.86959) * (4.185E+32 m^3) ] / [ 1.328E+20 m^3/s^2 ] } T = SQRT { [ (39.478) * (4.185E+32 m^3) ] / [ 1.328E+20 m^3/s^2 ] } T = SQRT { [ 1.652E+34 m^3 ] / [ 1.328E+20 m^3/s^2 ] } T = SQRT { 1.244E+14 s^2 } T = 11,155,579 s Which translates to: 129 days, 2 hours, 45 minutes and 36 seconds. Please mark as brainiest