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The torque on a planet about the cent...

The torque on a planet about the centre of sun is

A

Zero

B

Negative

C

Positive

D

Depend on mass of planet

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The correct Answer is:
To find the torque on a planet about the center of the Sun, we can follow these steps: ### Step 1: Understand the Definition of Torque Torque (τ) is defined as the cross product of the position vector (r) and the force vector (F). Mathematically, it is given by: \[ \tau = \mathbf{r} \times \mathbf{F} \] This can also be expressed in terms of the magnitudes of the vectors and the angle (θ) between them: \[ \tau = rF \sin(\theta) \] ### Step 2: Identify the Position and Force Vectors - The position vector (r) points from the center of the Sun to the planet. - The gravitational force (F) acting on the planet due to the Sun is directed towards the center of the Sun. ### Step 3: Determine the Angle Between the Vectors Since the force vector (F) is directed towards the Sun, and the position vector (r) points away from the Sun, the angle (θ) between these two vectors is 180 degrees. ### Step 4: Calculate the Sine of the Angle Using the sine function: \[ \sin(180^\circ) = 0 \] ### Step 5: Substitute into the Torque Formula Substituting the values into the torque formula: \[ \tau = rF \sin(180^\circ) = rF \cdot 0 = 0 \] ### Conclusion The torque on a planet about the center of the Sun is zero. ### Final Answer \[ \text{The torque on a planet about the center of the Sun is } 0. \] ---
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AAKASH INSTITUTE-GRAVITATION -ASSIGNMENT SECTION - A (OBJECTIVE TYPE QUESTIONS)
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  12. If the radius of the earth shrinks by 1.5% ( mass remaining same) , th...

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  13. If the density of the planet is double that of the earth and the radiu...

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  14. At what height above the surface of earth the value of "g" decrease...

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  15. During motion of a man from a equator to pole of earth , its wei...

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  16. If earth suddenly stop rotating , then the weight of an object of m...

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  17. If R is the radius of the earth and g the acceleration due to gravity ...

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  18. If the value of g at the surface of the earth is 9.8 m//sec^(2), then ...

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