Edexcel A Level (IAL) Physics-5.32 Newton’s Law of Universal Gravitation- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -5.32 Newton’s Law of Universal Gravitation- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -5.32 Newton’s Law of Universal Gravitation- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
Newton’s Law of Universal Gravitation
Newton’s law of universal gravitation states that every mass attracts every other mass with a gravitational force.
The Gravitational Force Equation
The magnitude of the gravitational force between two point masses is given by:
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\( F = \dfrac{G m_1 m_2}{r^2} \)
- \( F \) = gravitational force between the masses (N)
- \( G \) = gravitational constant \( = 6.67 \times 10^{-11}\ \mathrm{N\,m^2\,kg^{-2}} \)
- \( m_1, m_2 \) = masses (kg)
- \( r \) = distance between the centres of the masses (m)
Meaning of the Equation
- The force is proportional to the product of the two masses.
- The force is inversely proportional to the square of the distance between them.
- The force acts along the line joining the centres of the two masses.
- The force is always attractive.
Key idea: This is an inverse-square law.
Direction of the Gravitational Force
- Each mass experiences a force.
- The forces are equal in magnitude and opposite in direction.
- This is consistent with Newton’s third law of motion.
Using the Equation in Calculations
- Convert all values to SI units.
- Use the distance between the centres of the masses.
- Substitute carefully and calculate.
Common rearrangements:
\( m_1 = \dfrac{F r^2}{G m_2} \)
\( r = \sqrt{\dfrac{G m_1 m_2}{F}} \)
Applications of Newton’s Law of Gravitation
- Calculating gravitational attraction between planets.
- Explaining planetary motion and orbits.
- Deriving gravitational field strength around a planet.
- Understanding satellite motion.
Example (Easy)
Two masses of \( 2.0\,\mathrm{kg} \) and \( 3.0\,\mathrm{kg} \) are \( 0.50\,\mathrm{m} \) apart. Calculate the gravitational force between them.
▶️ Answer / Explanation
\( F = \dfrac{(6.67\times10^{-11})(2.0)(3.0)}{(0.50)^2} \)
\( F = 3.20\times10^{-9}\,\mathrm{N} \)
Example (Medium)
Calculate the gravitational force between the Earth \( (6.0\times10^{24}\,\mathrm{kg}) \) and a satellite of mass \( 500\,\mathrm{kg} \) orbiting at a distance of \( 7.0\times10^{6}\,\mathrm{m} \) from the Earth’s centre.
▶️ Answer / Explanation
\( F = \dfrac{(6.67\times10^{-11})(6.0\times10^{24})(500)}{(7.0\times10^{6})^2} \)
\( F \approx 4.1\times10^{3}\,\mathrm{N} \)
Example (Hard)
Two identical masses attract each other with force \( F \). If the distance between them is doubled, what is the new force?
▶️ Answer / Explanation
The force follows an inverse-square law:
\( F_{\text{new}} = \dfrac{F}{4} \)
