What is the binding energy of the earth-sun system neglecting the effect of other planets and satellites? (Mass of earth Me=6 × 1024 kg, mass of the sun Ms=2 × 1030 kg -2, Distance between earth sum =1.5 × 1011 m ) Assume circular orbit of the earth around the sun.

Question

What is the binding energy of the earth-sun system neglecting the effect of other planets and satellites? (Mass of earth Me=6 × 1024 kg, mass of the sun Ms=2 × 1030 kg -2, Distance between earth sum =1.5 × 1011 m ) Assume circular orbit of the earth around the sun. (A) 8.8 × 1010 J (B) 8.8 × 103 J (C) 5.2 × 1033 J (D) 2.6 × 1033 J

Tardigrade Admin · Accepted Answer

Binding energy of the system =(- G Me Ms/2 r) =(-6.6 × 10-11 × 6 × 1024 × 2 × 1030/2 × 1.5 × 1011) =-2.6 × 1033 J | BE |=2.6 × 1033 J

$8.8 \times 1010 J$

$8.8 \times 103 J$

$5.2 \times 1033 J$

$2.6 \times 1033 J$

Binding energy of the system $= \frac{- G M _{e} M _{s}}{2 r}$
$= \frac{- 6.6 \times 1 0 ^{- 11} \times 6 \times 1 0 ^{24} \times 2 \times 1 0 ^{30}}{2 \times 1.5 \times 1 0 ^{11}}$
$= - 2.6 \times 1 0^{33} J$
$∣ BE ∣ = 2.6 \times 1 0^{33} J$

Q. What is the binding energy of the earth-sun system neglecting the effect of other planets and satellites? (Mass of earth Me​=6×1024kg, mass of the sun Ms​=2×1030kg−2, Distance between earth & sum =1.5×1011m ) Assume circular orbit of the earth around the sun.

8.8×1010J

8.8×103J

5.2×1033J

2.6×1033J

Binding energy of the system =2r−GMe​Ms​​ =2×1.5×1011−6.6×10−11×6×1024×2×1030​ =−2.6×1033J ∣BE∣=2.6×1033J

$8.8 \times 1010 J$

$8.8 \times 103 J$

$5.2 \times 1033 J$

$2.6 \times 1033 J$

Binding energy of the system $= \frac{- G M _{e} M _{s}}{2 r}$
$= \frac{- 6.6 \times 1 0 ^{- 11} \times 6 \times 1 0 ^{24} \times 2 \times 1 0 ^{30}}{2 \times 1.5 \times 1 0 ^{11}}$
$= - 2.6 \times 1 0^{33} J$
$∣ BE ∣ = 2.6 \times 1 0^{33} J$