Question

Let S be an imaginary closed surface enclosing mass m. Let be an element of area on S, the direction of being outward from S. Let$\vec { E }$ be the gravitational intensity at . We define φ= , the integration being carried out over the entire surface S.

• A. φ = - Gm
• B. φ = - 4πGm
• C. φ = - Gm/4π
• D. No relation of the type (2), (2) or (3) can exist.

Answer 1

Answer: (B)

φ = - 4πGm

Answer 2

Follow the method used to prove Gauss’s law.

E = G.

= E dS cos (180^o- θ) = - EdS cos θ

φ = = - Gm. = -Gm.dω = -4πGm.