Two particles of masses ‘m’ and ‘9m’ are separated by a distance ‘r’. At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is (G = Universal constant of gravitation)

Question

Two particles of masses ‘m’ and ‘9m’ are separated by a distance ‘r’. At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is (G = Universal constant of gravitation) (A) - (4Gm/r) (B) - (8Gm/r) (C) - (16Gm/r) (D) - (32Gm/r)

Tardigrade Admin · Accepted Answer

Let electric field at point P is zero which is situated at a distance x from mass m. ∴ (G m/x2)=(G(9 m)/(r-x)2) Or ((r-x)2/x2)=9 Or ((r-x)/x)=3 ⇒ x=(r/4) Potential at point P V p =-( Gm / x )-( G (9 m )/ r - x ) Or Vp=-(G m/r / 4)-(G(9 m)/3 r / 4) Or V p =-( Gm / r )[4+12]=(-16 Gm / r )

Q. Two particles of masses ‘ $m$ ’ and ‘ $9 m$ ’ are separated by a distance ‘ $r$ ’. At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is ( $G =$ Universal constant of gravitation)

$- \frac{4 G m}{r}$

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

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

$- \frac{32 G m}{r}$

Q. Two particles of masses ‘m’ and ‘9m’ are separated by a distance ‘r’. At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is (G= Universal constant of gravitation)

−r4Gm​

−r8Gm​

−r16Gm​

−r32Gm​

Let electric field at point P is zero which is situated at a distance x from mass m. ∴x2Gm​=(r−x)2G(9m)​ Or x2(r−x)2​=9 Or x(r−x)​=3 ⇒x=4r​ Potential at point PVp​=−xGm​−r−xG(9m)​ Or Vp​=−r/4Gm​−3r/4G(9m)​ Or Vp​=−rGm​[4+12]=r−16Gm​

Q. Two particles of masses ‘ $m$ ’ and ‘ $9 m$ ’ are separated by a distance ‘ $r$ ’. At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is ( $G =$ Universal constant of gravitation)

$- \frac{4 G m}{r}$

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

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

$- \frac{32 G m}{r}$