Answer to Question #201906 in Mechanics | Relativity for fexypotato

Question

Two identical uniform spheres A and B, with mass M = 1. 00 [kg], are

placed 0. 500 [m] away from another spherical mass C (m =

0. 300 [kg]) as shown. If mass C is released from rest, what is its net

acceleration? Assume that only the gravitational forces from spheres A

and B act on C.

Answer 1

Mass of identical sphere(M)=1 kg

Mass of another sphere (m)=0.3 kg

Distance between sphere(d)=0.50 m

From the above figure,

Let us calculate the angle C by the law of cosines:

$cosC=\frac{AC^2+BC^2-AB^2}{2 \times AC \times BC} \\ = \frac{0.5^2+0.5^2-0.8^2}{2 \times 0.5\times 0.5} = -0.28$

Force due to A on C is given as:

$F_{AC}=\frac{Gm_Am_C}{AC^2}=\frac{GMm}{0.5^2}=F$

Force due to B on C is given as:

$F_{BC}=\frac{Gm_Am_C}{BC^2}= \frac{GMm}{0.5^2}=F$

The resultant or net force on C is given as:

$F_{net}= \sqrt{F_{AC}^2+F_{BC}^2+2F_{AC}F_{BC}cosC} \\ F_{AC}=F_{BC}=F \\ F_{net}=F\sqrt{2+2cosC} \\ =\frac{GMm}{d^2}\sqrt{2+2cosC} \\ = \frac{GMm}{0.5^2}\sqrt{2+2(-0.28)} = 4.8 \;Gmm$

Acceleration

$a= \frac{F_{net}}{m} \\ = \frac{4.8 \;Gmm}{m}= 4.8\;GM$

G = gravitational constant

$G= 6.67 \times 10^{-11} \;N m^2/kg^2 \\ a = 4.8 \times 6.67 \times 10^{-11} \times 1 = 32.03 \times 10^{-11} \;m/s^2$

Answer: $32.03 \times 10^{-11} \;m/s^2$