Newton’s Law of Universal Gravitation

by • 04/07/2012 • GeneralComments (0)420

Newton’s Law of Universal Gravitation

Consider tow bodies A and B having masses mA and mB respectively.

 

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Let,

F(AB) = Force on A by B

F(BA) = Force on B by A

r(AB) = displacement from A to B

r(BA) = displacement from B to A

r(AB) = unit vector in the direction of r(AB).

r(BA) = unit vector in the direction of r(BA).
From a(m) / g = Re2 / Rm2, we have

F(AB) ∞ 1 / r(BA)2 ……………………. (1)

Also,

F(AB) ∞ m(A) …………………………. (2)

F(BA) ∞ m(B)

According to theNewton’s third law of motion

F(AB) = F(BA) ……………….. (for magnitudes)

Therefore,

F(AB) ∞ m(B) ………………………….. (3)

Combining (1), (2) and (3), we get

F(AB) ∞ m(A)m(B) / r(BA)2

F(AB) = G m(A)m(B) / r(BA)2 ……………………… (G = 6.67 x 10(-11) N – m2 / kg2)
Vector Form

F(AB) = – (G m(A)m(B) / r(BA)2) r(BA)

F(BA) = – (G m(B)m(A) / r(AB)22) r(AB)

Negative sign indicates that gravitational force is attractive.
Statement of the Law

“Every body in the universe attracts every other body with a force which is directly proportional to the products of their masses and inversely proportional to the square of the distance between their centres.”

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