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Question: In the solar system, the sun is in the focus of the system for the sun-earth binding system. Then th...

In the solar system, the sun is in the focus of the system for the sun-earth binding system. Then the binding energy for the system will be (given that radius of the earth’s orbit around the sun is 1.5×1011m1.5 \times {10^{11}}m and mass of the earth =6×1024Kg6 \times {10^{24}}Kg)
A) 2.7×1033J2.7 \times {10^{33}}J
B) 5.4×1033J5.4 \times {10^{33}}J
C) 2.7×1030J2.7 \times {10^{30}}J
D) 5.4×1030J5.4 \times {10^{30}}J

Explanation

Solution

In our solar system every planet is bound within the effect of the sun. This is due to the law of gravity by the sun on each planet in the solar system. That’s why our earth keeps on staying in its particular orbit. If the velocity of the earth is much larger than its present velocity it can escape from its orbit. The energy keeps the earth in a particular orbit is called gravitational potential energy.

Formula used:
U=GMmRU = \dfrac{{GMm}}{R}
Where,
G= Gravitational constant which equals to 6.67×1011Nm2kg26.67 \times {10^{11}}N{m^2}k{g^{ - 2}}
M= Mass of the sun
M= Mass of the earth
R= distance between the center of the earth and the sun

Complete step by step answer:
(i) From the question, we can understand that the binding energy between the sun and the earth is the gravity exerted by the sun on earth. And it keeps the earth at particular orbit. Hence energy is called gravitational potential energy. The gravitational potential energy,
U=GMmRU = \dfrac{{GMm}}{R}
(ii) Applying the given values in the above formula,
U=6.67×1011×1030×6×10241.5×1011\Rightarrow U = \dfrac{{6.67 \times {{10}^{11}} \times {{10}^{30}} \times 6 \times {{10}^{24}}}}{{1.5 \times {{10}^{11}}}}
We can add and subtract the power terms as the base number is the same. We get the equation as,
U=6.67×61.5×1011+30+2411\Rightarrow U = \dfrac{{6.67 \times 6}}{{1.5}} \times {10^{ - 11 + 30 + 24 - 11}}
By simplifying the given equation, we get,
U=26.68×1032\Rightarrow U = 26.68 \times {10^{32}}
We can consider the approximate value,
U2.7×1033J\therefore U \approx 2.7 \times {10^{33}}J

Therefore, the correct option is A.

Note:
The gravitational force is one of the fundamental forces in nature. This is due to the distance between the objects which are acting under gravity are separated at large distances. The gravitational potential will be zero if the point where the energy is the surface of the earth. This is called a zero point.