Question

Two bodies of masses 5 & 50 kg respectively, are separated by a distance 10 . Where

should an object of mass 5 be placed between them so that the gravitational force acting on this

object is zero?​

Answer 1

Answer:

Explanation:

Now gravitational force on middle body due to left body = gravitational force on middle body due to right body so that the overall gravitational force becomes zero.

$\frac{G(5)(5)}{x^{2}} = \frac{G(50)(5)}{(10-x)^{2}}\\\\(10-x)^{2} = 10x^{2}\\\\(10-x)^{2} - (x\sqrt{10})^{2} = 0\\\\(10 - x + x\sqrt{10})(10-x-x\sqrt{10}) = 0\\\\10 = x - x\sqrt{10}\\\\or\\\\10 = x+x\sqrt{10}\\\\x = \frac{10}{1-\sqrt{10}}\\\\or\\\\x = \frac{10}{1+\sqrt{10}}\\\\now \\\\x = \frac{10}{1+\sqrt{10}}$

because x cannot be negative solve to get the value of x as 2.4m (approx)

Thus the 5kg mass must be placed at a distance of 2.4m from 5kg mass

Thanks! Might help!