4.2 Relate gravitational force, field strength, mass, and distance | University of Alberta EN PH 131: Engineering Mechanics: Dynamics

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

Gravitation and Orbital Motion

EN PH 131 study topic 4.2

Gravity is an attractive interaction between masses. To describe it, we need to know which objects we are considering, how far apart they are, and which direction the force acts. In this lesson, the system is a chosen test object, and a source mass is the object producing the gravitational field. We use an observer at rest in an inertial reference frame, with a fixed coordinate system; unless stated otherwise, take the positive radial direction to point away from the source. The source attracts the test object, so the gravitational force points opposite that positive direction. We will relate force, field strength, mass, and separation using Newton's law of universal gravitation.

What you will learn

Relate the gravitational force between two masses to their masses and separation.
Explain gravitational field strength as force per unit mass and calculate it from a source mass and distance.
Distinguish mass from weight and use consistent SI units and directions.
Check gravitational calculations using dimensions and how force and field strength change with distance.

1. Force between two masses

For two objects treated as particles, let their masses be

m_1

and

m_2

, and let the distance between their centres be

r

. The gravitational force on either object points along the line joining the centres, toward the other object. The two forces have equal magnitudes and opposite directions.

Newton's law of universal gravitation gives the magnitude of either force. Here

G

is the universal gravitational constant. It is not the same for different planets or objects: it is a constant of nature. Use masses in kilograms and separation in metres to obtain force in newtons.

A useful model is that a spherical body outside its surface produces the same gravitational field as if all its mass were concentrated at its centre. Thus, for an object above a planet,

r

is measured from the planet's centre, not from the surface. This relation applies when the spherical-body model is appropriate.

F=G\frac{m_1m_2}{r^2}

The force is attractive and acts along the line between the two masses.
Doubling either mass doubles the force; doubling the separation reduces the force to one quarter.
Distance in the law is centre-to-centre distance.

2. Field strength and the force on a test mass

Gravitational field strength describes the gravitational influence at a location. It is defined as the gravitational force on a small test mass divided by that test mass. The test mass is useful for measuring the field, but its mass cancels when the field is calculated from a source mass.

For a source mass

M

and a location a distance

r

from its centre, the field-strength magnitude decreases with the square of the distance. Its direction is toward the source. In vector form, if the outward radial unit vector is

\hat{\mathbf r}

, then the field vector has a negative sign because it points inward.

A test mass

m

placed in a field experiences force equal to its mass multiplied by the field vector. Near Earth's surface, field strength is commonly approximated as

9.81\ \mathrm{N/kg}

, corresponding to an acceleration of about

9.81\ \mathrm{m/s^2}

for freely falling objects when other forces are negligible. This is an approximation for locations close to the surface.

\mathbf g=-\frac{GM}{r^2}\hat{\mathbf r},\qquad \mathbf F=m\mathbf g

Field strength is force per unit mass, measured in newtons per kilogram.
For a given source, field strength depends on source mass and distance, not on the test mass.
Weight is the gravitational force on an object at a specified location; mass is a property of the object.

3. Choosing distance, direction, and checks

Before calculating, identify the source mass and the object experiencing the force. Choose a reference frame and coordinates that make the line between them clear. For a radial calculation, define the outward direction as positive; gravity then has a negative radial component. If only magnitudes are requested, use positive magnitudes and state the direction in words.

For an object near a spherical planet, use

r=R+h

, where

R

is the planet's radius and

h

is height above its surface. Substituting height alone would put the object the wrong distance from the planet's centre. The local field can be found from

g=GM/r^2

, and then the object's weight magnitude is

F=mg

.

Dimensions provide a quick check:

G

has units of

\mathrm{N\,m^2/kg^2}

, so multiplying by two masses and dividing by distance squared leaves newtons. Dividing force by mass gives

\mathrm{N/kg}

, which is equivalent to

\mathrm{m/s^2}

. As distance becomes very large, the field and force approach zero. As source or test mass increases, force increases in direct proportion; increasing source mass increases field strength.

r=R+h,\qquad g=\frac{GM}{r^2}

Use centre-to-centre distance and convert all quantities to SI units.
Keep vector direction separate from magnitude.
Use the inverse-square dependence to check whether a result changes plausibly.

Worked example

Weight at Earth's surface

Estimate the gravitational force on a 70.0 kg student standing at Earth's surface using Earth's mass

M_E=5.972\times10^{24}\ \mathrm{kg}

and radius

R_E=6.371\times10^6\ \mathrm{m}

. Use

G=6.67430\times10^{-11}\ \mathrm{N\,m^2/kg^2}

. Take the student as the system and the Earth's centre as the source.

Student attracted toward Earth

The weight points toward Earth's centre, opposite the positive outward direction.

Set the frame and knowns
Use a frame fixed to Earth for this force calculation, with the radial axis positive outward. The student is the system; Earth is the source. The force direction is inward. The known quantities are the two masses, Earth's radius, and $G$ ; the unknown is the force magnitude.
Calculate the field
At the surface, the centre-to-centre distance is Earth's radius. Calculate the field magnitude first, so the role of the student's mass remains clear.
$g=\frac{GM_E}{R_E^2}=9.82\ \mathrm{N/kg}$
Calculate the force
Multiply the field magnitude by the student's mass. The vector points inward, so its radial component is negative under the chosen sign convention.
$F=mg=(70.0\ \mathrm{kg})(9.82\ \mathrm{N/kg})=687\ \mathrm{N}$

Answer: The gravitational force has magnitude approximately

687\ \mathrm{N}

and points toward Earth's centre.

Check: The field is close to the expected near-surface value, and kilograms multiplied by newtons per kilogram gives newtons. The inward direction agrees with attraction.

Worked example

Field strength above a planet

A small probe is at altitude

h=4.00\times10^5\ \mathrm{m}

above Earth. Estimate Earth's field strength there and the gravitational force on a

1.20\times10^3\ \mathrm{kg}

probe. Use the same Earth data as in Example 1. Treat Earth as spherical and neglect other bodies.

Probe above Earth's surface

The force points inward; the radius used in the model starts at Earth's centre.

Define the geometry
The probe is the system, Earth is the source, and the observer uses a frame fixed to Earth's centre. Choose outward as positive radial direction. The initial and final states are not needed because this is a force calculation, not a motion calculation. The distance to the source centre is $r=R_E+h$ .
$r=6.371\times10^6+4.00\times10^5=6.771\times10^6\ \mathrm{m}$
Find the field strength
Apply the spherical-source field relation at the probe's distance. Its direction is inward, while the value below is the magnitude.
$g=\frac{GM_E}{r^2}=8.69\ \mathrm{N/kg}$
Find the probe's force
The probe's mass multiplies the field magnitude. Report the direction separately to avoid confusing the positive magnitude with the signed radial component.
$F=mg=(1.20\times10^3\ \mathrm{kg})(8.69\ \mathrm{N/kg})=1.04\times10^4\ \mathrm{N}$

Answer: The field strength is approximately

8.69\ \mathrm{N/kg}

inward, and the force magnitude on the probe is approximately

1.04\times10^4\ \mathrm{N}

inward.

Check: The altitude increases the centre distance, so the field is smaller than the near-surface value. The force units reduce to newtons.

Worked example

Force between two separated masses

Two compact objects with masses

8.00\ \mathrm{kg}

and

3.00\ \mathrm{kg}

have centres separated by

0.500\ \mathrm{m}

. Find the magnitude and direction of the force on each object. Use

G=6.67430\times10^{-11}\ \mathrm{N\,m^2/kg^2}

.

Mutual gravitational attraction

The paired forces are equal in magnitude and opposite in direction.

Set the system and direction
Consider the two objects in one inertial frame, with the positive horizontal axis from the 8.00 kg object toward the 3.00 kg object. The separation is the centre-to-centre distance. The unknown is the force magnitude; each force points toward the other object.
Apply the two-mass law
Use both masses and the given separation in the inverse-square law. The same calculated magnitude applies to the force on each object.
$F=G\frac{m_1m_2}{r^2}=6.41\times10^{-9}\ \mathrm{N}$
State the directions and verify
The force on the 8.00 kg object is positive along the chosen axis, and the force on the 3.00 kg object is negative along it. Their equal magnitudes and opposite directions are consistent with the mutual attraction between the objects.
$\mathbf F_{8\to3}=-\mathbf F_{3\to8}$

Answer: Each object experiences a gravitational force of magnitude

6.41\times10^{-9}\ \mathrm{N}

, directed toward the other object.

Check: The result has force units because

Gm_1m_2/r^2

reduces to newtons. Increasing the separation would reduce the force.

Common mistakes and how to avoid them

Using altitude above a planet's surface as the distance in the inverse-square law.

Correction: Measure

r

from the planet's centre. For an object at altitude

h

, use

r=R+h

.

Treating field strength and gravitational force as the same quantity.

Correction: Field strength is force per unit mass; multiply it by the test object's mass to obtain the force.

Confusing mass in kilograms with weight in newtons.

Correction: Mass is the object's amount of matter in this model; weight is its gravitational force at a particular location.

Reporting a positive radial force when the chosen positive direction is outward.

Correction: Gravity points inward. State the direction clearly, or use a negative radial component with outward defined as positive.

Lesson summary

Two masses separated by centre-to-centre distance $r$ attract with magnitude $F=Gm_1m_2/r^2$ .
A source mass $M$ produces field strength magnitude $g=GM/r^2$ , directed toward the source.
A test mass in a field experiences force $\mathbf F=m\mathbf g$ .
For a spherical planet at altitude $h$ , use the centre distance $r=R+h$ .
Check units, direction, and inverse-square behaviour before accepting a result.

Check your understanding

Question 1

For a fixed source mass, what happens to gravitational field strength if the distance from its centre is doubled?

It doubles.
It becomes one half as large.
It becomes one quarter as large.
It stays the same.

Show answer and explanation

It becomes one quarter as large.

Field strength varies as

1/r^2

. Replacing

r

by

2r

divides the field by

2^2=4

.

Question 2

A test mass is tripled while its location in a fixed gravitational field stays the same. What happens to the force magnitude?

It is divided by three.
It stays the same.
It triples.
It increases by a factor of nine.

Show answer and explanation

It triples.

Since

F=mg

and the field is unchanged, tripling

m

triples the force.

Key terms

Gravitational force: The attractive force between masses, directed along the line joining them.
Gravitational field strength: Gravitational force per unit test mass at a location; its SI unit is newtons per kilogram.
Test mass: The object whose force is used to describe the gravitational field at a location.
Weight: The gravitational force acting on an object at its location.
Centre-to-centre distance: The separation between the centres of two objects in the particle or spherical-source model.

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4.2 · Relate gravitational force, field strength, mass, and distance

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 study topic 4.2

What you will learn

1. Force between two masses

2. Field strength and the force on a test mass

3. Choosing distance, direction, and checks

Weight at Earth's surface

Field strength above a planet

Force between two separated masses

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through EN PH 131

About this lesson and its review