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Three identical point mass each of mass ...

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

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`

Text Solution

Verified by Experts

The correct Answer is:
C

`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)`
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Knowledge Check

  • The centre of mass of three bodies each of mass 1 kg located at the points (0,0), (3, 0) and (0, 4) in the x-y plane is

    A
    `((4)/(3) , 1)`
    B
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    D
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    A
    `Gm^(2)`
    B
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