Question

Determine the small gravitational force F which the copper sphere exerts on the steel sphere. Both spheres are homogeneous, and the value of r is 65 mm. Express your result as a vector.

Answer 1

Answer:

$F=3.665 \times 10^{-7} N$ acting along the line joining both the sphere and always attractive in nature.

Explanation:

The given radius of both the sphere, r= 65 mm = 0.065 m

So, the volume of the spheres, $v= \frac 4 3 \pi r^3$

$v= \frac 4 3 \pi (0.065)^3 = 1.150 \times 10^{-3} m^3$

The density of steel, $\rho _s = 7850 kg/m^3$

and the density of copper, $\rho_c= 8940 kg/m^3$

Let M be the mass of the copper ball and m  is the mass of the steel ball.

So, $M=\rho_c v= 8940\times 1.150 \times 10^{-3} = 10.281$ kg

$m=\rho_s v= 7850\times 1.150 \times 10^{-3} = 9.0275$ kg

The gravitational force, F, between the two objects having masses M and m and separated by distance d is

$F=\frac{GMm}{d^2}$

Where $G= 6.674 30 x 10^{-11} m^3 kg^{-1} s^{-2}$ is the universal gravitational constant.

When both the sphere touches each other, d = 2r= 2 x 0.065 = 0.13 m

Hence, the gravitational force between both the sphere,

$F= \frac {6.674 30 x 10^{-11}\times 10.281 \times 9.0275}{0.13^2} \\\\F=3.665 \times 10^{-7} N$

The nature of gravitational force is always attractive and acting along the line joining the center of both the sphere.