Question

What happened to the force of gravity between two object if
(1) distance between object triple
(2) mass of both object double
(3) mass of both as well as distance between them double

Answer 1

HEY MATE!

YOUR ANSWER...

Formula of force of gravity is F= G Mm/(d)^2

So, 1) when distance between the object is triple,
F=G Mm/(3d)^2
F= G Mm/9d^2
F=1/9G Mm/d^2
So, force becomes 1/9 times.

2) When mass of both objects is doubled,
F=G 2(Mm)/d^2
F=G 2M2m/d^2
F=G 4Mm/d^2
F=4G Mm/d^2
So, force becomes 4 times.

3) When mass of both objects as well as distance between them is doubled,
F=G 2(Mm)/(2d)^2
F=G 4Mm/4d^2
F=4/4 G Mm/d^2
F=G Mm/d^2
So, force becomes 1 time.

HOPE IT HELPS!

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Answer 2

Your answer is ----


according to formula of force of gravity

F = G (Mm)/r^2

where ,

F = force of gravity b/w two objectes

G = gravitational constant

M , m = mass to two objects

r = distance between two objects


F is directly proportional to the product of mass of two objects

also, F is inversly proportional to the square of distance between two objects


So,

(i) when distance between two object become triple , force of gravity become 1/9 of it's original value

(ii) when mass of both object become double , force of gravity become 4 time of it's original value

(iii) when mass of both as well as distance b/w them become double , then it's force of gravity become same .


Mathematically :-

F = G(Mm)/r^2 .....(1)

(i) when distance become double

F' = G(Mm)/(3r)^2

=> F' = GMm/9r^2

=> F' = 1/9 F (from 1)


(ii) when both masses become double

F' = G(2M2m)/r^2

=> F' = 4 F


(iii) when mass of both as well as distance between them double

F' = G(2M2m)/(2r)^2

=> F' = 4GMm/4r^2

=> F' = GMm/r^2

=> F' = F