Question

the gravitational force between the two objects is 49N.How much distance between these objects be decreased so that force becomes double?
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Answer 1

GmM / r2

100 = Gm M/ r2

r2= GmM/100

now for the second case

F' = GmM / r'250 = GmM / r'2

r'2= GmM / 50

or

r'2= 2 r2

thus,

r' = r√2

Answer 2

Given,

The gravitational force between two objects = 49 N

To find,

The amount by which the distance between the two objects be decreased so that the force becomes double.

Solution,

We can simply solve this numerical problem by using the following process:

Let us assume that the mass of both the objects is M and m. And, let the initial and the final distance of separation between the two objects be r and R, respectively.

As per gravitational law;

The gravitational force acting between two bodies of mass M and m, separated by a distance d, is mathematically represented as;

F = (G ×M×m)/R^2,

where G = Gravitational constant

= 6.67408 × 10-11 m3 kg-1 s-2

M = mass of the first object

m = mass of the second object

R = distance of separation

between the two objects

Now, according to the question;

When the initial distance of separation between the two objects is r, then the gravitational force equals 49 N

=> G × (M×m)/r^2 = 49 N {Equation-1}

Now,

When the new distance of separation between the two objects is R, then the gravitational force is doubled

=> G × (M×m)/R^2 = 2 × 49 N

=> G × (M×m)/R^2 = 2 × G × (M×m)/r^2

G × (M×m)/R^2 = 2 × G × (M×m)/r^2 (according to equation-1)

=> 1/R^2 = 2/r^2

=> R^2 = r^2/2

=> R = r/√2

=> the final distance of separation between the two objects is 1/√2 times the initial distance of separation

Hence, the final distance of separation between the two objects is 1/√2 times the initial distance of separation.