Answer:
0.00373 N
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
[tex]m_1=m_2=m_3=7700\ kg[/tex] = Mass of the boxes
[tex]r_1[/tex] = Distance of one box from origin = 100 cm
[tex]r_1[/tex] = Distance of other box from origin = 420 cm
From the universal law of gravity
[tex]F_1=\frac{Gm_1m_2}{r_1^2}\\\Rightarrow F_1=\frac{Gm^2}{r_1^2}[/tex]
On the right hand side
[tex]F_2=\frac{Gm^2}{r_2^2}[/tex]
The force on the center block will be
[tex]F=F_1-F_2\\\Rightarrow F=\frac{Gm^2}{r_1^2}-\frac{Gm^2}{r_2^2}\\\Rightarrow F=Gm^2\left(\frac{1}{r_1^2}-\frac{1}{r_2^2}\right)\\\Rightarrow F=6.67\times 10^{-11}\times 7700^2\left(\frac{1}{1^2}-\frac{1}{4.2^2}\right)\\\Rightarrow F=0.00373\ N[/tex]
The magnitude of the net gravitational force on the mass at the origin due to the other two masses is 0.00373 N
The magnitude of the net gravitational force Fgrav on the mass at the origin due to the other two masses is mathematically given as
F= 0.00373N
Question Parameter(s):
Three identical very dense masses of 7700kgeach
One mass is at x1= -100cm
one is at the origin
one is at x2 = 420cm
Generally, the equation for the universal law of gravity is mathematically given as
[tex]F_1=\frac{Gm_1m_2}{r_1^2} \\\\\\Rightarrow F_1=\frac{Gm^2}{r_1^2}[/tex]
[tex]F_2=\frac{Gm^2}{r_2^2}[/tex]
therefore
[tex]F=F_1-F_2\\\\\Rightarrow F=\frac{Gm^2}{r_1^2}-\frac{Gm^2}{r_2^2}\\\\\Rightarrow F=Gm^2\left(\frac{1}{r_1^2}-\frac{1}{r_2^2}\right)\\\\\Rightarrow F=6.67\times 10^{-11}\times 7700^2\left(\frac{1}{1^2}-\frac{1}{4.2^2}\right)\\\\\Rightarrow F=0.00373\ N[/tex]
In conclusion magnitude of the net gravitational force is
F= 0.00373N
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