What is the force of the circular motion of the moon?

The moon (mass= 7.36 x 10^22 kg) orbits the earth at a range of 3.84 x 10^5 km with a period of 28 days. Determine force of circular motion of the moon.What is the force of the circular motion of the moon?The force comes from the gravitational attraction to the earth. It can be calculated by m*v^2/r. v can be calculated by v=d/t=circumference/period.



v=2*pi*3.84x10^8m/(28days*24h*3600 s)=997m/s



F=mv^2/r=7.36x10^22kg*(997m/s)^2/3.84x鈥?br>


=1.9 x 10^20 NWhat is the force of the circular motion of the moon?w^2 R = a = g is the gravity field in N/kg on the Moon at radius R from Earth. w = 2pi/T is its angular velocity with T = 28*24*60*60 seconds as its period.



Thus the centripetal force F = mg = mw^2 R = 7.36E22*(2pi/(28*24*60*60))^2*3.84E8 = 1.90645E+20 Newtons. m is the mass of the moon.



Note a is the centrifugal acceleration that is offset by g the gravitational field when an orbit is in a fixed radius R. Thus we have w^2 R = a = g so that the force F = mg.