# I Problem with gravitation field perpendicular vector.

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1. Nov 22, 2018

### Abdu Ewais

since it is known that $\vec{A_\perp} = -{mG \over R^2}$ why did the professor write it as $\vec{A_\perp} = {- R G \rho \over 3}$ for perfect sphere with perfect mass distribution ? Shouldn't it be $\vec{A_\perp} = -{4 \over 3} \pi R G \rho$? I need help thanks.

Last edited: Nov 22, 2018
2. Nov 22, 2018

### Staff: Mentor

If I understand your question, shouldn't that be a 4/3 instead of a 4? (But I agree that your professor's version seems off.)

3. Nov 22, 2018

### Abdu Ewais

Sorry got it off by a bit there. fixed