Question

Three masses of 1kg, 2kg, and 3kg, are placed at the vertices of an equilateral traingle of side 1m. Find the gravitational potential energy of this system. Take G= 6.67xx10^(-11) N-m^(2)//kg^(2).

Answer 1

Explanation:

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Answer 2

Answer:

The gravitational potential energy is 7.337* 10^-10 Nm

Explanation:

The gravitational potential energy of two massages is given by

(GMm)\R;

where M= mass of first body

m=mass of second body

R= distance between the two

bodies

Here we will have to calculate the sum of the gravitational potential energy of the corresponding pairs of bodies.

Therefore,

Total gravitational potential energy=

{G*1*2\1 + G*2*3\1 + G*1*3\1}Nm

=G*(2+6+3)Nm

=( 6.67*10^-11)* 11 NM

=7.337* 10^-10 Nm