Gravitational potential energy of a system of particles as shown in the figure is

Question

Gravitational potential energy of a system of particles as shown in the figure is (A) (G m1 m2/r1)+(G m2 m3/r3)+(G m1 m3/r3) (B) ((-G m1 m2/r1))+((-G m2 m3/r2))+((-G m1 m3/r3)) (C) (-G m1 m2/r1)-(G m2 m3/r2)+(G m1 m3/r3) (D) (G m1 m2/r1)+(G m2 m3/r2)-(G m1 m3/r3)

Tardigrade Admin · Accepted Answer

For a system of particles, all possible pairs are taken and total gravitational potential energy is the algebraic sum of the potential energies due to each pair, applying the principle of superposition.
Total gravitational potential energy

= \frac{- G m _{1} m _{2}}{r _{1}} - \frac{G m _{2} m _{3}}{r _{2}} - \frac{G m _{1} m _{3}}{r _{3}}

= (\frac{- G m _{1} m _{2}}{r _{1}}) + (\frac{- G m _{2} m _{3}}{r _{2}}) + (\frac{- G m _{1} m _{3}}{r _{3}})

Q. Gravitational potential energy of a system of particles as shown in the figure is

$\frac{G m _{1} m _{2}}{r _{1}} + \frac{G m _{2} m _{3}}{r _{3}} + \frac{G m _{1} m _{3}}{r _{3}}$

$(\frac{- G m _{1} m _{2}}{r _{1}}) + (\frac{- G m _{2} m _{3}}{r _{2}}) + (\frac{- G m _{1} m _{3}}{r _{3}})$

$\frac{- G m _{1} m _{2}}{r _{1}} - \frac{G m _{2} m _{3}}{r _{2}} + \frac{G m _{1} m _{3}}{r _{3}}$

$\frac{G m _{1} m _{2}}{r _{1}} + \frac{G m _{2} m _{3}}{r _{2}} - \frac{G m _{1} m _{3}}{r _{3}}$

Q. Gravitational potential energy of a system of particles as shown in the figure is

r1​Gm1​m2​​+r3​Gm2​m3​​+r3​Gm1​m3​​

(r1​−Gm1​m2​​)+(r2​−Gm2​m3​​)+(r3​−Gm1​m3​​)

r1​−Gm1​m2​​−r2​Gm2​m3​​+r3​Gm1​m3​​

r1​Gm1​m2​​+r2​Gm2​m3​​−r3​Gm1​m3​​

$\frac{G m _{1} m _{2}}{r _{1}} + \frac{G m _{2} m _{3}}{r _{3}} + \frac{G m _{1} m _{3}}{r _{3}}$

$(\frac{- G m _{1} m _{2}}{r _{1}}) + (\frac{- G m _{2} m _{3}}{r _{2}}) + (\frac{- G m _{1} m _{3}}{r _{3}})$

$\frac{- G m _{1} m _{2}}{r _{1}} - \frac{G m _{2} m _{3}}{r _{2}} + \frac{G m _{1} m _{3}}{r _{3}}$

$\frac{G m _{1} m _{2}}{r _{1}} + \frac{G m _{2} m _{3}}{r _{2}} - \frac{G m _{1} m _{3}}{r _{3}}$