Correct Answer - Option 3 :
\(\frac{4}{3} \pi G (dm x)\)
The correct answer is option 3) i.e. \(\frac{4}{3} \pi G (dm x)\)
CONCEPT:
-
Law of Universal Gravitation: It states that all objects attract each other with a force that is proportional to the masses of two objects and inversely proportional to the square of the distance that separates their centres.
It is given mathematically as follows:
\(F = \frac{Gm_1m_2}{R^2}\)
Where m1 and m2 are the mass of two objects, G is the gravitational constant and R is the distance between their centres.
- From the Law of Universal Gravitation, the gravitational force (F) acting on an object of mass m placed on the surface of Earth is
\(F = \frac{GMm}{R^2}\)
Where R is the radius of the earth and M is the mass of the Earth.
EXPLANATION:
The gravitational force acting on the body if kept on the surface of Earth = F = \( \frac{GMm}{R^2}\)
- Let M' be the effective mass of Earth for a distance x < R.
Let the mean density of the earth be 'd' and the radius of Earth corresponding to the effective mass of Earth is 'x'.
The effective mass of Earth = Mean density × Effective volume
⇒ \(M' = d \times \frac{4}{3} \pi x^3\) (∵ volume of sphere = \(\frac{4}{3} \pi r^3\))
Gravitational force, \( F = \frac{GM'm}{x^2} = \frac{G(d \times \frac{4}{3} \pi x^3)m}{x^2}=\frac{4}{3} \pi G (dm x)\)
