Question

You place a 0.17 kg can of soup and a 0.31 kg jar of pickles on the kitchen counter, separated by a distance of 0.42 m. What is the magnitude of the force of gravity that the can of soup exerts on the jar of pickles?

Answer 1

$1.984 \times 10^{-11} \mathrm{N} \text { is the force of gravity exerted on the jar of pickles. }$

Explanation:

According to Newton's third law that each force has an equal and opposite reaction force in this case both of the jars will exert the same force an each other

. The force is given by

$\mathrm{F}=\frac{G \times M_{1} \times M_{2}}{r^{2}}$

Where, F = force, $G=\text { gravitational constant }=\left(6.67 \times 10^{-11}\right)$, $mass \left(\mathrm{M}_{1}\right)=0.17 \mathrm{kg}$, $mass \left M_{2}= 0.31 \mathrm{kg}$ and Distance(r) = 0.42 m.

Substitute the values in the formula.

$\mathrm{F}=\frac{6.67 \times 10^{-11} \times 0.17 \times 0.31}{0.42^{2}}$

$\mathrm{F}=\frac{3.51 \times 10^{-12}}{0.176}$

$\mathrm{F}=1.984 \times 10^{-11} \mathrm{N}$

$\text { The force of gravity exerted on the jar of pickles is } 1.984 \times 10^{-11} \mathrm{N} \text { . }$

Answer 2

The magnitude of the force of gravity that the can of soup exerts on the jar of pickles is 2.0 x 10⁻¹¹ N.

The given parameters;

• mass of the first object, m₁ = 0.17 kg
• mass of the second object, m₂ = 0.31 kg
• distance between the two object, r = 0.42 m

The magnitude of the force of gravity that the can of soup exerts on the jar of pickles is calculated by using Newton's law of universal gravitation as follows;

$F = \frac{Gm_1m_2 }{r^2} \\\\F = \frac{6.67 \times 10^{-11} \times 0.17\times 0.31}{0.42^2} \\\\F = 2 .0 \times 10^{-11} \ N$

Thus, the magnitude of the force of gravity that the can of soup exerts on the jar of pickles is 2.0 x 10⁻¹¹ N.

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