Answer to Question #100155 in Field Theory for Mahi

Question #100155
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
1
Expert's answer
2019-12-10T09:55:30-0500

Assume m1=2×109 kgm_1 = 2×10^{-9} \text{ kg}, m2=25×109 kgm_2 = 25×10^{-9} \text{ kg}

1) Distances to masses r=r1=r2=3 mr = r_1 =r_2=3 \text{ m}



Gravitational field, created by masses:


g1=F1m=Gm1r2,g2=F2m=Gm2r2g_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


g=(g2g1)ex=G(m2m1)r2ex\vec{g} = (g_2-g_1)\vec{e_x} = \frac{G(m_2-m_1)}{r^2}\vec{e_x}

g=G(m2m1)r2=1.8541019 ms2|\vec{g}| = \frac{G(m_2-m_1)}{r^2} = 1.854*10^{-19} \text{ ms}^{-2}

2) Distances to masses r=r1=r2=32+42 m=5 mr = 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 : α=2arccos45\alpha = 2 \arccos\frac{4}{5} , cosα=725\cos \alpha = \frac{7}{25}

Hence,

g=g12+g222g1g2cos(πα)=Gr2m12+m22+2m1m2cos(2arccos45)|\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})}

g=Gr2m12+m22+1425m1m2=6.8431020 ms2|\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}


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