Gravitational potential energy of the mass at the centre of a square as shown in figure is Potential energy at infinite is taken to be zero

Question

Gravitational potential energy of the mass at the centre of a square as shown in figure is Potential energy at infinite is taken to be zero (A) - (4Gm2/l√3) (B) - (2√2 Gm2/l) (C) - (-4 Gm2/l) (D) - (4√2 Gm2/l)

Tardigrade Admin · Accepted Answer

U = - Δ V Ã m = (Vo-V∞)m = (-Gm×4× m/(l)√2)=-(4√2Gmm/l)=-(4√2Gm2/l)

Q. Gravitational potential energy of the mass at the centre of a square as shown in figure is {Potential energy at infinite is taken to be zero}

$- \frac{4 G m ^{2}}{l 3}$

$- \frac{2 2 G m ^{2}}{l}$

$- \frac{- 4 G m ^{2}}{l}$

$- \frac{4 2 G m ^{2}}{l}$

U = $- Δ V \times m$
= $(V_{o} - V_{\infty}) m$
$= \frac{- G m \times 4 \times m}{\frac{l}{2}} = - \frac{4 2 G mm}{l} = - \frac{4 2 G m ^{2}}{l}$

Q. Gravitational potential energy of the mass at the centre of a square as shown in figure is {Potential energy at infinite is taken to be zero}

−l3​4Gm2​

−l22​Gm2​

−l−4Gm2​

−l42​Gm2​

U = −ΔV×m = (Vo​−V∞​)m =2​l​−Gm×4×m​=−l42Gmm​​=−l42​Gm2​

$- \frac{4 G m ^{2}}{l 3}$

$- \frac{2 2 G m ^{2}}{l}$

$- \frac{- 4 G m ^{2}}{l}$

$- \frac{4 2 G m ^{2}}{l}$

U = $- Δ V \times m$
= $(V_{o} - V_{\infty}) m$
$= \frac{- G m \times 4 \times m}{\frac{l}{2}} = - \frac{4 2 G mm}{l} = - \frac{4 2 G m ^{2}}{l}$