Question

Four point masses each of mass $m$ are placed on vertices of a regular tetrahedron. Distance between any two masses is $r$

• A. Gravitation field at centre is zero
• B. Gravitation potential energy of system in $\frac{-4G{m}^{}}{r}$
• C. Gravitational potential energy of system in $\frac{-6G{m}^2}{r}$
• D. Gravitation force on one of the point mass is $\frac{\sqrt{6}G{m}^2}{r^2}$

Answer 1

Answer: (A), (C)

Gravitation field at centre is zero
Gravitational potential energy of system in $\frac{-6G{m}^2}{r}$

We know that,

Tetrahedron is the symmetric structure.

According to the figure

The equal masses are placed at every corner point of the tetrahedron.

So,

$m_1=m_2=m_3=m_4=m$

The centre of the tetrahedron will lie equidistant from all the four vertices.

Thus,

$r_1=r_2=r_3=r_4$

Therefore,

$F_{12}=F_{23}=F_{31}=F_{14}=F_{24}=F_{34}$

At the centre of the tetrahedron, due to symmetry, the net gravitational field will be zero.

Gravitational Potential Energy:

The gravitational potential energy of the system will be given by: $\begin{array}{cc}& -(\frac{G{m}^2}{r_{12}}+\frac{G{m}^2}{r_{23}}+\frac{G{m}^2}{r_{34}}+\frac{G{m}^2}{r_{13}}+\frac{G{m}^2}{r_{14}}+\frac{G{m}^2}{r_{24}})\\ & =-\frac{6G{m}^2}{r}\end{array}$