Three identical point mass each of mass 1kg lie in the x-y plane at point (0,0), (0,0.2m) and (0.2m, 0). The net gravitational force on the mass at the origin is

Options

A. `1.67xx10^(-11)(hati+hatj)N`
B. `3.34xx10^(-10)(hati+hatj)N`
C. `1.67xx10^(-9)(hati+hatj)N`
D. `3.34xx10^(-10)(hati-hatj)N`

Answer

C

Explanation

`F_(1)=F_(2)=(G(1)(1))/((.2)^(2))=(6.67xx10^(-11))/(.04)=1.67xx10^(-9)`
`vecF_("net")=F_(1)(hati)+F_(2)(hatj)=F(hati+hatj)=1.67xx10^(-9)(hati+hatj)`