Answer to Question #159925 in Classical Mechanics for Hillary

Question

A 200kg object and a 500kg object are separated by 4.00m Find
(a)the net gravitational force exerted by these objects on a 50.0 kg object placed midway between them.
(b) At position (other than an infinitely remote one) can the 50.0kg object be placed so as to experience a net force of zero from the other two objects?

Answer 1

(a) Let’s calculate the resultant gravitational force on a mass $m=50\ kg$. According to the Newton’s law of universal gravitation, the mass $m_1=200\ kg$ attracts the mass $m$ with a gravitational force $F_1$. The object $m_2=500\ kg$ attracts the mass $m$ with a gravitational force $F_2$. Then, the sum of these two forces, taken algebraically (because they are along the same line) will be the net gravitational force $F_{net}$:

$$F_{net}=F_1+F_2,$$ $$F_{net}=-G\dfrac{m_1m}{r^2}+G\dfrac{m_2m}{r^2},$$ $$F_{net}=\dfrac{Gm}{r^2}(m_2-m_1),$$ $$F_{net}=\dfrac{6.67\cdot10^{-11}\ \dfrac{Nm^2}{kg^2}\cdot50\ kg}{(2\ m)^2}\cdot(500\ kg-200\ kg)=2.5\cdot10^{-7}\ N.$$

(b) At a point between the two objects at a distance $d$ from the $500\ kg$ object, the net force on the $50\ kg$ object will be zero when $F_1=F_2$:

$$G\dfrac{m_1m}{(r-d)^2}=G\dfrac{m_2m}{d^2},$$ $$\dfrac{m_1}{(r-d)^2}=\dfrac{m_2}{d^2},$$ $$\dfrac{d}{r-d}=\sqrt{\dfrac{m_2}{m_1}},$$ $$d=\dfrac{r\sqrt{\dfrac{m_2}{m_1}}}{1+\sqrt{\dfrac{m_2}{m_1}}},$$ $$d=\dfrac{4\ m\cdot\sqrt{\dfrac{500\ kg}{200\ kg}}}{1+\sqrt{\dfrac{500\ kg}{200\ kg}}}=2.45\ m.$$