
The earth (mass = \[6 \times {10^24}\]kg) revolves around the sun with an angular velocity of \[{\text{2}} \times {\text{1}}{{\text{0}}^{ - 7}}\] rad/s in a circular orbit of radius \[1.5 \times {10^8}\]. The force exerted by the sun on the Earth, in Newtons, is:
(A) \[36 \times {10^{21}}\]
(B) \[27 \times {10^{39}}\]
(C) 0
(D) \[18 \times {10^{25}}\]
Answer
232.8k+ views
Hint The force exerted by the sun on earth is the same as force exerted by earth on sun which is the consequence of Newton’s 3rd law of motion.
Complete step-by-step solution
When an object is moving in a circular path, it experiences a force which is called a centripetal force. This centripetal force pulls the object towards the center of the circle at all times. This centripetal force is given by
\[F = m{\omega ^2}r\]
Where F= centripetal force
M=mass of the moving object
ω= angular velocity of the object
r= distance of the object from the axis of rotation.
Substituting the values given in question,
M=\[6 \times {10^{24}}\]kg
W=\[{\text{2}} \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}\] rad/s
R=\[1.5 \times {10^8}\]km
F= \[6 \times {10^{24}}\] \[ \times {\text{ (2}} \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}{)^2}\] \[ \times {\text{ 1}}{\text{.5}} \times {\text{1}}{{\text{0}}^8}\]
F= \[36 \times {10^{21}}\]N
Therefore, the correct answer is option A.
Note This force is centripetal force experienced by earth and it is balanced by the force of gravity between the Earth and the Sun. If there was no centripetal force, Earth would have been pulled inside the sun long ago.
Complete step-by-step solution
When an object is moving in a circular path, it experiences a force which is called a centripetal force. This centripetal force pulls the object towards the center of the circle at all times. This centripetal force is given by
\[F = m{\omega ^2}r\]
Where F= centripetal force
M=mass of the moving object
ω= angular velocity of the object
r= distance of the object from the axis of rotation.
Substituting the values given in question,
M=\[6 \times {10^{24}}\]kg
W=\[{\text{2}} \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}\] rad/s
R=\[1.5 \times {10^8}\]km
F= \[6 \times {10^{24}}\] \[ \times {\text{ (2}} \times {\text{1}}{{\text{0}}^{{\text{ - 7}}}}{)^2}\] \[ \times {\text{ 1}}{\text{.5}} \times {\text{1}}{{\text{0}}^8}\]
F= \[36 \times {10^{21}}\]N
Therefore, the correct answer is option A.
Note This force is centripetal force experienced by earth and it is balanced by the force of gravity between the Earth and the Sun. If there was no centripetal force, Earth would have been pulled inside the sun long ago.
Recently Updated Pages
Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

