Two equal masses separated by a distance d attract each other with a force (F). If one unit of mass is transferred from one of them to the Other, the force

Question

Two equal masses separated by a distance d attract each other with a force (F). If one unit of mass is transferred from one of them to the Other, the force (A) does not change (B) decreases by (G/d2) (C) becomes d2 times (D) increases by (2G/d2)

Tardigrade Admin · Accepted Answer

F ( text initial) =( text G.m.m / d 2) F ( text final )=( text G. ( m +1) ⋅( m -1)/ d 2)=( G ⋅( m 2-1)/ d 2) Clearly from above expressions we have a decrease in force by ( G / d 2).

Q. Two equal masses separated by a distance d attract each other with a force (F). If one unit of mass is transferred from one of them to the Other, the force

does not change

decreases by $(G / d^{2})$

becomes $d^{2}$ times

increases by $(2 G / d^{2})$

$F_{(initial)} = \frac{G.m.m}{d ^{2}} F_{(final)} = \frac{G. ( m + 1 ) \cdot ( m - 1 )}{d ^{2}} = \frac{G \cdot ( m ^{2} - 1 )}{d ^{2}}$
Clearly from above expressions we have a decrease in force by $\frac{G}{d ^{2}}$ .

Q. Two equal masses separated by a distance d attract each other with a force (F). If one unit of mass is transferred from one of them to the Other, the force

does not change

decreases by (G/d2)

becomes d2 times

increases by (2G/d2)

F(initial) ​=d2 G.m.m ​F(final )​=d2 G. (m+1)⋅(m−1)​=d2G⋅(m2−1)​ Clearly from above expressions we have a decrease in force by d2G​.

decreases by $(G / d^{2})$

becomes $d^{2}$ times

increases by $(2 G / d^{2})$

$F_{(initial)} = \frac{G.m.m}{d ^{2}} F_{(final)} = \frac{G. ( m + 1 ) \cdot ( m - 1 )}{d ^{2}} = \frac{G \cdot ( m ^{2} - 1 )}{d ^{2}}$
Clearly from above expressions we have a decrease in force by $\frac{G}{d ^{2}}$ .