13.1–13.3 - Gravitational forces and fields

13.1–13.3 - Gravitational forces and fields

Use the gravitational field model to describe forces on test masses, apply Newton's inverse-square law, and derive field strength outside a spherical mass. Read field directions and graphs, and explain why a small region near a planet's surface can be treated as approximately uniform.

A field maps force per unit mass

Does a planet's gravitational field disappear when no satellite happens to be there? No. The field describes what force a small test mass would experience at each position; the test mass helps define or measure it, rather than switching it on.

A mass is the source of a gravitational field. The field is a model mapping the gravitational force per unit mass through space. It is not material emitted by the source. At a particular point, a larger test mass experiences a proportionally larger force, while the force per unit mass is unchanged.

Gravitational field strength

Gravitational field strength at a point is the gravitational force per unit mass on a small test mass placed at that point.

The field is a vector: its direction is the direction of force on a test mass. Gravitational forces are attractive, so the field around an isolated spherical planet points towards its centre.

Force per unit mass

g=FmF=mgg=\frac{F}{m} \qquad F=mg

Here gg is field-strength magnitude in N kg−1\mathrm{N\,kg^{-1}}, FF is gravitational force in N\mathrm{N}, and mm is test mass in kg\mathrm{kg}. The force and field vectors have the same direction. The test mass is small enough not to significantly alter the source arrangement.

Not on the data sheet: learn it.

The unit N kg−1\mathrm{N\,kg^{-1}} is equivalent to m s−2\mathrm{m\,s^{-2}}, since a newton is kg m s−2\mathrm{kg\,m\,s^{-2}}. Thus field strength is also the acceleration of a freely falling test mass when gravity is its only force. A supported object can experience a gravitational force without accelerating because another force balances it.

[DIAGRAM: asset_name: Gravitational field lines are a map; asset_slug: gravitational-field-map; caption: Inward radial lines map the exterior field; a small surface region can be approximated by parallel lines; description: The left panel shows a spherical mass with twelve non-crossing radial lines outside it and arrows pointing inward. The right panel enlarges a small near-surface patch and shows approximately parallel evenly spaced downward arrows. Labels state that lines represent field direction and relative strength, not emitted material.]
Inward radial lines map the exterior field

Field-line arrows show direction. Closer line spacing represents greater strength in the chosen drawing; lines do not cross because the field at one point has one resultant direction. The number of lines an artist chooses is not a count of physical objects. A small surface patch is often drawn with parallel lines as a local approximation, not as a claim that the whole planet's field is parallel.

Watch out

Mass is not field strength. Doubling a test mass doubles its gravitational force, but does not double the field strength at that position. Distinguish the source, the field and the test mass.

Newton's law and spherical masses

Every pair of masses attracts gravitationally. Increasing either mass increases the force; increasing their separation weakens it. Newton's law gives the quantitative relation for point masses.

Newton's law of gravitation

F=Gm1m2r2F=\frac{Gm_1m_2}{r^{2}}

Here FF is force magnitude in N\mathrm{N}, masses m1,m2m_1,m_2 are in kg\mathrm{kg}, and rr is their separation in m\mathrm{m}. The forces act along the line joining the masses, towards one another. The gravitational constant is G=6.67×10−11 N m2 kg−2G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}} using the examination value.

The value of GG is on the data sheet. The force equation is not on the data sheet: learn it.

The inverse square is decisive: doubling separation reduces the force to one quarter. Doubling one mass doubles the force; doubling both masses multiplies it by four. The force on each body has the same magnitude, even when their masses are very different, by Newton's third law. Their accelerations can differ because acceleration is force divided by the body's own mass.

Worked example 1: the attraction of laboratory masses

Question. Two small masses of 8.0 kg8.0\,\mathrm{kg} and 12 kg12\,\mathrm{kg} have centres 0.30 m0.30\,\mathrm{m} apart. Treat them as point masses and calculate their gravitational attraction using G=6.67×10−11 N m2 kg−2G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}.

Plan. Use the centre separation and both masses.

F=6.67×10−11×8.0×120.302 N=7.11×10−8 NF=\frac{6.67\times10^{-11}\times8.0\times12}{0.30^{2}}\,\mathrm{N} =7.11\times10^{-8}\,\mathrm{N}

Answer. Approximately 7.1×10−8 N7.1\times10^{-8}\,\mathrm{N} on each mass, in opposite directions towards one another.

Check. Ordinary laboratory masses have extremely small mutual gravitational forces. Do not confuse this force with either mass's much larger weight in Earth's field.

For a point outside a uniform sphere, its gravitational effect is the same as if the sphere's mass were concentrated at its centre. This justifies using a centre-to-centre distance for spherical planets. A height above the surface is not that distance: r=R+hr=R+h, where RR is planet radius and hh altitude.

Exam-style example: a force outside a model planet

Question. A spherical planet has mass 6.0×1024 kg6.0\times10^{24}\,\mathrm{kg}. A 2.0×103 kg2.0\times10^{3}\,\mathrm{kg} spacecraft is 7.0×106 m7.0\times10^{6}\,\mathrm{m} from its centre. Calculate the gravitational force, using G=6.67×10−11 N m2 kg−2G=6.67\times10^{-11}\,\mathrm{N\,m^{2}\,kg^{-2}}.

Plan. Treat the exterior spherical planet as a point mass at its centre.

F=6.67×10−11×6.0×1024×2.0×103(7.0×106)2 N=1.63×104 NF=\frac{6.67\times10^{-11}\times6.0\times10^{24}\times2.0\times10^{3}} {(7.0\times10^{6})^{2}}\,\mathrm{N} =1.63\times10^{4}\,\mathrm{N}

Answer. Approximately 1.6×104 N1.6\times10^{4}\,\mathrm{N}, towards the planet's centre.

Check. The separation is squared as a whole, including its power of ten. The point-mass representation applies outside the sphere, not automatically inside it.

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