Calculate the radius of the orbit of Mars as it circles the sun in units of m.
Given for purposes of this problem: Mars (mass of 2 x 10^30 kg) requires 2.46 years to orbit the sun, which has a an almost circular trajectory.
The gravitational constant is 6.672 x 10^-11 N m^2/kg^2.
2006-11-17 06:01:57 · 2 answers · asked by Dee 4 in Science & Mathematics ➔ Physics
Kepler's 3rd law says T = (2*pi*R^1.5)/sqrt(G*M) where M is the mass of the central body (i.e., the sun, not Mars) ~ 2*10^30 kg. Then R = [sqrt(G*M)*T/(2*pi)]^(2/3). I get 2.7298*10^11 m.
2006-11-17 07:04:15 · answer #1 · answered by kirchwey 7 · 0⤊ 0⤋
F=G*M*m/r^2; F=m*a; a=r*w^2; F is centripetal force, a is centripetal acceleration, w is angular speed of Mars, M=2*10^30kg mass of the sun (not Mars), m is mass of Mars, r is the distance btw M & m. Hence G*M*m/r^2=m*r*w^2, hence r=(G*M/w/w)^(1/3), whereas angular speed of Mars w=2pi/(2.46years*365.25days*24hours*3600s). Thus r=2.731*10^11m –Yes!
2006-11-17 18:58:10 · answer #2 · answered by Anonymous · 0⤊ 0⤋