Answer :
Answer:
[tex]g=8.97\times 10^{-11}\ N/kg[/tex]
Explanation:
It is given that,
Mass of three objects, [tex]m_1=m_2=m_3=3.4\ kg[/tex]
Edge of square, l = 2.2 m
Gravitational constant, [tex]G=6.67259\times 10^{-11}\ Nm^2/kg^2[/tex]
Let g is the magnitude of the gravitational field at the fourth corner due to these masses. It is equal to the sum of magnitude due to diagonal mass and magnitudes due to two neighborhood masses.
[tex]g=\dfrac{Gm}{(\sqrt{2l} )^2}+\dfrac{Gm\sqrt{2} }{l^2}[/tex]
[tex]g=\dfrac{Gm}{l^2}(\dfrac{1}{2}+\sqrt{2})[/tex]
[tex]g=\dfrac{6.67259\times 10^{-11}\times 3.4}{(2.2)^2}(\dfrac{1}{2}+\sqrt{2})[/tex]
[tex]g=8.97\times 10^{-11}\ N/kg[/tex]
So, the magnitude of the gravitational field at the fourth corner due to these masses is [tex]8.97\times 10^{-11}\ N/kg[/tex]. Hence, this is the required solution.