physics question need help!!!! + need the work or explination for it?

Question

calculate the centripetal acceleration oh the earth in its orbit around the sun,and the net force exerted on the earth. What exerts this force on the Earth? Assume that the Earth's orbit is a circle of radious 1.50 x10 to the 11th.

Answer 1

Acceleration
a = w^2 r
w = 2pi / T
w = 2 x 3.14 / (365.25x24x60x60)
w = 1.99 x10^ - 7
a = (1.99 x10^ - 7) ^2 x 1.5 x 10 ^11
a = 5.95 x 10 ^ -3 m/s^2

Force

F = mw^2r
F = 6.02 x 10^24 x (1.99 x10^ - 7) ^2 x 1.5x10^11
F = 3.58 x 10^ - 22 N

Source of this force:
It is the gravitational force of attraction between the sun and the
Earth. It is given by Newton's law of gravitation.

Answer 2

OK; the Earth's orbit is a circle radius 1.5x10^11 metres. Centripetal acceleration is (v^2)/r. To work out v, the distance travelled by the Earth in a year is one circle; 2x pi x 1.5x10^11, which my calculator says is 9.42x10^11. To turn this into metres/sec, divide by the number of seconds in a year; you get 29871 metres/sec. Square this and divide by r, as in the formula, and you get 0.0059 metres/ sec squared.This is the acceleration due to the Sun's gravity which holds the Earth in its orbit and prevents it shooting off into space in a straight line. If you stopped the Earth in its orbit and allowed it to drop into the Sun, it would accelerate towards the Sun at this rate, which is only about 1/200 g. That's a very gentle acceleration, but in about 3 months the Earth would have fallen 150 million km. into the Sun.