# Two balls, each with a mass of 0.813 kg, exert a gravitational force of 8.38 x 10^-11 N on each other.How far apart are the balls? The value of the universal gravitational constant is 6.673 x...

Two balls, each with a mass of 0.813 kg, exert a gravitational force of 8.38 x 10^-11 N on each other.

How far apart are the balls? The value of the universal gravitational constant is 6.673 x 10^-11 N m^2/kg^2. Answer in units of m

Asked on by jayy000

pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted on

The formula for gravitational attraction is

Fg = (Gm1m2)/r^2

In this formula, G is the Universal Gravitational Constant while m1 and m2 are the masses of the two bodies.  R is the distance of separation between the centers of the two masses.

In the example you have provided, we have the gravitational force already given and we have the universal constant and the masses of the two bodies.  This gives us:

### 8.38 x 10^-11 = (6.673 x 10^-11*.813*.813)/r^2

Now this is just algebra.  We multiply both sides by r^2.

That gets us

### 8.38 x 10^-11*r^2 = 4.45 x 10^-11

We divide to find r^2 and get

r^2 = .530503

by taking the square root we get

r = .728356

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Let the distance between the two balls be d and their masses be m1 and m2. Then by gravitational law, the force of attraction F,between them is:

F =G*m1*m2/d^2 .

By data,F = 8.38 *10^(-11)N,  m1 = m2 = 0.813 kg , G =6.673 x 10^-11 N m^2/kg^2 and d is to be determined.

Therefore,

d =sqrt( G*m1*m2/F ). Substituting the values from data,

d = sqrt{(6.673 x 10^-11)(0.813)^2/(8.38*10^-11)}

=0.725486104 m is the required distance between the two masses, where the gravitational attractional   force  between them  is as given.

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