Question

Two bodies of masses 5kg and 15kg are separated by a distance of 5m.Calculate the gravitational force between them.​

Answer 1

F = (G × m × M )/ d² = (G × 5×15)/25 N = 75G/25 N

Now, as you know the value of G = 6.7 × 10^(-11), so,

F = (6.7 × 10^(-11) × 75)/25 N = 20.1 × 10^(-11) N = 20 × 10^(-11) N [Approx.]

The sign ^ means to the power.

Answer 2

ANSWER:

$\textsf{Force between the two bodies} \ \sf =2 \times 10^{-10} \ N$

GIVEN:

• Two bodies of masses 5 kg and 15 kg are separated by a distance of 5 m.

TO FIND:

• Gavitational force between the two bodies.

EXPLANATION:

$\boxed{ \bold{ \gray{ \large{F=\dfrac{Gm_1m_2}{d^2}}}}}$

Here G is the universal gravitational constant.

$\sf G = 6.673 \times 10^{-11}\ Nm^2kg^{-2}$

$\sf m_1 = 5 \ kg$

$\sf m_2 = 15\ kg$

$\sf d = 5\ m$

$\sf F =\dfrac{6.673 \times 10^{-11} \times 5(15)}{5^2}$

$\sf F =6.673 \times 10^{-11} \times 3$

$\sf F =20 \times 10^{-11}$

$\sf F =2 \times 10^{-10} \ N$

Hence the force between the two bodies = $\bf 2 \times 10^{-10} \ N$

NEWTON's LAW OF GRAVAITATION:

Newton's law of gravitation states that every object in the universe attracts every other object with a force that is directly proportional to the product of their massess and inversely proportinal to the square of the distance between them.

Diagram:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0) \thicklines\put( 0, 0){\line(1,0){4}}\put( - 0.6, - 0.1){ \bf m_1 }\put(4.1, - 0.1){ \bf m_2 }\multiput(0, 0)(4, 0){2}{\circle*{0.1}} }\put( 0, - 0.1){ \underbrace{ \qquad\qquad\qquad\qquad\qquad \: \: \: \: \: \: } }\put(1.85, - 0.65){ \bf d }\end{picture}$

Derivation:

$\sf F \propto m_1 m_2$

$\sf F \propto \dfrac{1}{d^2}$

$\sf F \propto \dfrac{m_1 m_2}{d^2}$

$\sf F = \dfrac{Gm_1 m_2}{d^2}$

Here G is the constant of propotionality and is called universal gravitational constant.