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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.

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Edexcel A level Physics-Study Notes- All Topics

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:

\( F = \dfrac{G m_1 m_2}{r^2} \)

  • \( F \) = 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:

\( F_{\text{new}} = \dfrac{F}{4} \)

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