Answer to Question #100155 in Field Theory for Mahi

Question

In a rectangular coordinate system, a most of 2×10^-9 kg is a placed at the origin of the coordinates and a most of 25×10^-9 kg is at the point of x=6m and y=0m.

What is the gravitational field at 1.x=3m,y=0m 2.x=3m,y=4m

Answer 1

Assume $m_1 = 2×10^{-9} \text{ kg}$, $m_2 = 25×10^{-9} \text{ kg}$

1) Distances to masses $r = r_1 =r_2=3 \text{ m}$

Gravitational field, created by masses:

$$g_1 = \frac{F_1}{m} = \frac{Gm_1}{r^2},\, g_2 = \frac{F_2}{m} = \frac{Gm_2}{r^2}$$

As gravitational field is Vector sum of gravitational field gives

$$\vec{g} = (g_2-g_1)\vec{e_x} = \frac{G(m_2-m_1)}{r^2}\vec{e_x}$$

$$|\vec{g}| = \frac{G(m_2-m_1)}{r^2} = 1.854*10^{-19} \text{ ms}^{-2}$$

2) Distances to masses $r = r_1 =r_2=\sqrt{3^2+4^2} \text{ m} = 5 \text{ m}$

Use a law of cosine to find resulting magnitude:

angle between forces F₁ and F₂ : $\alpha = 2 \arccos\frac{4}{5}$ , $\cos \alpha = \frac{7}{25}$

Hence,

$|\vec{g}| = \sqrt{g_1^2+g_2^2-2 g_1 g_2 \cos (\pi - \alpha)} = \frac{G}{r^2}\sqrt{m_1^2+m_2^2+2 m_1 m_2 \cos (2 \arccos\frac{4}{5})}$

$|\vec{g}| = \frac{G}{r^2}\sqrt{m_1^2+m_2^2+\frac{14}{25} m_1 m_2 } = 6.843*10^{-20} \text{ ms}^{-2}$