Two bodies of masses m and 9 m are placed at a distance r. The gravitational potential at a point on the line joining them, where gravitational field is zero, is ( G is universal gravitational constant)

Question

Two bodies of masses m and 9 m are placed at a distance r. The gravitational potential at a point on the line joining them, where gravitational field is zero, is ( G is universal gravitational constant) (A) (-14 Gm /r) (B) (-16 Gm /r) (C) (-12 Gm /r) (D) (-8 Gm /r)

Tardigrade Admin · Accepted Answer

Mass of first body (m1)=m Mass of second body (m2)=9 m Distance between the bodies (d)=r Now, distance of the point from the body of mass m, where gravitational field is zero is given by x=(d/√(m2)m1+1) or, x=(r/√(9 m)m+1) or, x=(r/4) Now, (r-x)=r-(r/4)=(3 r/4) ∴ Potential, v=v1+v2 =-(G m1/x)-(G m2/(r-x)) =-(G m/r / 4)-(G(9 m)/3 r / 4) =-(4 G m/r)-(12 G m/r) =-(16 G m/r)

Q. Two bodies of masses $m$ and $9 m$ are placed at a distance $r$ . The gravitational potential at a point on the line joining them, where gravitational field is zero, is ( $G$ is universal gravitational constant)

$\frac{- 14 G m}{r}$

$\frac{- 16 G m}{r}$

$\frac{- 12 G m}{r}$

$\frac{- 8 G m}{r}$

Q. Two bodies of masses m and 9m are placed at a distance r. The gravitational potential at a point on the line joining them, where gravitational field is zero, is ( G is universal gravitational constant)

r−14Gm​

r−16Gm​

r−12Gm​

r−8Gm​

Q. Two bodies of masses $m$ and $9 m$ are placed at a distance $r$ . The gravitational potential at a point on the line joining them, where gravitational field is zero, is ( $G$ is universal gravitational constant)

$\frac{- 14 G m}{r}$

$\frac{- 16 G m}{r}$

$\frac{- 12 G m}{r}$

$\frac{- 8 G m}{r}$