Find the net gravitational force on a $2.0\,\text{kg}$ sphere midway between a $4.0\,\text{kg}$ sphere and a $7.0\,\text{kg}$ sphere that are $1.2\,\text{m}$ apart.

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Jason · Accepted Answer

The distance from the midway sphere to either outer sphere is:$$ r = \dfrac{1.2\,\text{m}}{2} = 0.60\,\text{m} $$
The gravitational force exerted by the $4.0\,\text{kg}$ sphere on the $2.0\,\text{kg}$ sphere is directed toward the $4.0\,\text{kg}$ sphere:$$ F_{4 \to 2} = \dfrac{G m_4 m_2}{r^2} = \dfrac{(6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2)(4.0\,\text{kg})(2.0\,\text{kg})}{(0.60\,\text{m})^2} \approx 1.48 \times 10^{-9}\,\text{N} $$
The gravitational force exerted by the $7.0\,\text{kg}$ sphere on the $2.0\,\text{kg}$ sphere is directed toward the $7.0\,\text{kg}$ sphere:$$ F_{7 \to 2} = \dfrac{G m_7 m_2}{r^2} = \dfrac{(6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2)(7.0\,\text{kg})(2.0\,\text{kg})}{(0.60\,\text{m})^2} \approx 2.59 \times 10^{-9}\,\text{N} $$
Since the two forces act in opposite directions along the line connecting the centers of the spheres, the net gravitational force is:$$ F_{\text{net}} = F_{7 \to 2} &#8211; F_{4 \to 2} = 2.59 \times 10^{-9}\,\text{N} &#8211; 1.48 \times 10^{-9}\,\text{N} = 1.11 \times 10^{-9}\,\text{N} $$
or factoring:$$ F_{\text{net}} = \dfrac{G m_2 (m_7 &#8211; m_4)}{r^2} = \dfrac{(6.67 \times 10^{-11}\,\text{N}\cdot\text{m}^2/\text{kg}^2)(2.0\,\text{kg})(7.0\,\text{kg} &#8211; 4.0\,\text{kg})}{(0.60\,\text{m})^2} \approx 1.11 \times 10^{-9}\,\text{N} $$

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