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13.2.1Circular motion and gravitational fields (A Level)

Point Mass Approximation

5 min read 7 parts Lesson 6 of 14 1 quiz

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Lesson summary

13.2.1 Point Mass Approximation — AS PHYSICS 9702 notes: Use the point mass approximation for uniform spheres and state Newton’s law of gravitation.Uniform sphere — for a point outside a uniform sphere, its mass can be considered concentrated at a single point at its centre of ma…

  • Field lines
  • Orbiting bodies
  • Relationship
  • Doubling the distance
  • Geostationary
  • Inverse square law

Section 1 of 7Point mass approximation

Uniform sphere
  • Uniform sphere for a point outside a uniform sphere, its mass can be considered concentrated at a single point at its centre of mass.
Uniform
  • Uniform the mass is distributed evenly.
Field lines
  • Field lines those around a uniform sphere are identical to those around a point mass.
Condition
  • Condition an object can be regarded as a point mass when it has an even mass distribution and the distance considered is larger than its size.
Planets
  • Planets this applies to the field lines around planets.

Section 2 of 7Newton’s law of gravitation

Statement
  • Statement the gravitational force between two point masses is proportional to the product of the masses and inversely proportional to the square of their separation.
Equation
  • FG = Gm1m2/r2.
  • FG is the gravitational force in N.
  • G is Newton’s gravitational constant.
  • m1 and m2 are the point masses in kg.
  • r is the distance between the centres of the masses in m.
Orbiting bodies
  • Orbiting bodies the law applies to planets orbiting the Sun, since stars and planets are approximately uniform spheres.

Section 3 of 7Inverse square law

Relationship
  • F is proportional to 1/r2.
Doubling the distance
  • Doubling the distance the force becomes a quarter of its value.
Distance
  • Distance r is measured from the centre of each mass.

Section 4 of 7Using the law

1
List the masses, the force and the distances.
2
Find r as the distance from the centre of the planet, which is the height above the surface plus the planet’s radius.
3
Rearrange FG = Gm1m2/r2 for the unknown, for example m2 = Fr2/Gm1.
4
Convert km to m and substitute.

Section 5 of 7More from the specification

1Spherical mass symmetry in gravitational fields
  • Spherical mass symmetry in gravitational fields homogeneous spherical bodies generate external radial gravitational fields identical to concentrated point masses.
2Newton's law of universal gravitation
  • Every point mass attracts every other point mass with a radial force: F = G M m / r2.
3Orbital speed in circular orbits
  • Equating gravitational force to centripetal requirement gives the orbital speed of a satellite: v = sqrt(GM / r).
4Geostationary orbit parameters
  • Geostationary satellites orbit directly above the equator with a 24-hour period, remaining fixed relative to a point on Earth's surface at an orbital radius of approximately 42,200 km.
5Newton's law of universal gravitation
  • Every point mass attracts every other point mass with a radial force: F = G M m / r2.
    What the symbols mean: F = gravitational force (N) · G = 6.67 × 10−11 N m2 kg−2 · m1, m2 or M, m = masses (kg) · r = distance between their centres (m)
6Orbital speed in circular orbits
  • Equating gravitational force to centripetal requirement gives the orbital speed of a satellite: v = sqrt(GM / r).
7Geostationary orbit parameters
  • Geostationary satellites orbit directly above the equator with a 24-hour period, remaining fixed relative to a point on Earth's surface at an orbital radius of approximately 42,200 km.
8Centripetal force
  • Net force toward the centre, mv2/r.
9Angular speed
  • Magnitude of the angular velocity.
10Gravitational field
  • Region where mass experiences gravitational force.
11Circular motion
  • Motion along a circular path with a force towards the centre.
12Instantaneous velocity
  • Velocity at one instant.
13Satellite
  • A body that orbits a larger one; the Moon is the Earth's natural satellite.
14Orbit
  • Path of one body around another under gravity.
15Geostationary
  • Orbiting with a period of 24 h above the equator, so staying above one point.
16Velocity
  • Rate of change of displacement.
17Period
  • Time for one revolution.
18Geostationary orbit
  • 24 h equatorial orbit above a fixed point on Earth.

Section 6 of 7Key terms

TermMeaning
Point massMass treated as concentrated at one point.
Uniform sphereSphere with its mass distributed evenly.
Newton’s law of gravitationF = Gm₁m₂/r² for point masses.
Inverse square lawForce is proportional to 1/r².
Gravitational constantConstant G in Newton’s law of gravitation.

Section 7 of 7Exam technique

1
Forgetting to add the planet’s radius to the height above the surface to find r.
2
Thinking the force is proportional to 1/r — it is 1/r².
3
Using the distance between the surfaces of two spheres instead of their centres.
4
Confusing the force on each mass — the two forces are equal in magnitude.
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