4.1 Apply Newton’s law of universal gravitation | 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 4.1

Any two objects with mass attract one another. Newton’s law of universal gravitation gives the force between them. In this lesson, the system is the object whose force we want to find; another mass acts as the source of that force. An observer describes both positions in a chosen Cartesian reference frame. The force calculation depends on the masses and their positions, not on whether either mass is moving. The law gives the force magnitude, while the positions determine its direction. A useful first step is to draw the line joining the masses and point the force toward the source mass.

What you will learn

State Newton’s law of universal gravitation and identify its quantities.
Use positions and coordinates to find the magnitude and direction of gravitational force.
Distinguish gravitational force from gravitational acceleration and weight.
Check gravitational calculations using units, direction, and limiting cases.

1. The gravitational-force model

For two particles with masses

m_1

and

m_2

separated by a distance

r

, the gravitational-force magnitude is proportional to the product of the masses and inversely proportional to the square of their separation. The universal gravitational constant is

G=6.6743\times10^{-11}\ \mathrm{N\,m^2/kg^2}

. Use kilograms and metres to obtain force in newtons.

Gravity is attractive. If mass 1 is the system and mass 2 is the source, the force on mass 1 points from mass 1 toward mass 2. The force on mass 2 has the same magnitude and opposite direction. These forces act on different objects, so they do not cancel on one object’s free-body diagram.

For a vector description, let

\mathbf r_1

and

\mathbf r_2

be the position vectors of the masses in the chosen frame. The vector from mass 1 to mass 2 is

\mathbf r_2-\mathbf r_1

. Dividing this vector by its length gives a unit vector in the force direction. When several source masses act, find each force and add the force vectors.

F=G\frac{m_1m_2}{r^2},\qquad \mathbf F_{1\leftarrow2}=Gm_1m_2\frac{\mathbf r_2-\mathbf r_1}{|\mathbf r_2-\mathbf r_1|^3}

Use straight-line, centre-to-centre separation in the point-mass model.
Gravitational force acts along the line joining the masses and points toward the source.
For several sources, add force vectors, not just force magnitudes.

2. Coordinates, signs, and gravitational acceleration

Before calculating, name the system and source, identify the observer’s reference frame, and choose axes with stated positive directions. Record each mass’s position in that frame. Subtract the system’s position from the source’s position to obtain the vector pointing toward the source. Reversing this subtraction reverses the direction.

For masses on one line, it is often simplest to calculate the force magnitude first and then assign a positive or negative component according to the chosen axis. For a two-dimensional arrangement, use components or the vector formula. Do not add magnitudes directly when the force directions differ.

Newton’s second law relates the net force on an object to its actual acceleration. Gravitational acceleration, by contrast, is the gravitational force per unit mass at a location. For a spherical body treated as a point mass outside its surface, the field magnitude at distance

r

from its centre is

g=GM/r^2

, directed toward the centre. Near Earth’s surface, this is commonly approximated as

9.81\ \mathrm{m/s^2}

; this is an approximation to the same gravitational interaction.

Mass is measured in kilograms; weight is a force measured in newtons. Near Earth’s surface, an object’s weight is approximately its mass times the local gravitational acceleration. When the distance from Earth’s centre changes, the gravitational acceleration changes according to the inverse-square relation.

g=\frac{GM}{r^2},\qquad \mathbf W=m\mathbf g

Choose positive directions before assigning component signs.
Gravitational acceleration is gravitational force per unit mass; it is not necessarily the object’s actual acceleration if other forces act.
For a spherical source outside its surface, measure distance from its centre.

3. A reliable solution and verification process

State the system and observer, then list the given masses and positions in SI units. Identify the unknown: force magnitude, force components, or gravitational acceleration. Draw a particle sketch when it helps show the source direction and axes. A sketch need not be to scale.

Find the separation. If coordinates are supplied, subtract the positions to form the separation vector and calculate its length. Apply the inverse-square law to the magnitude, or use its vector form to calculate components. If there are several sources, calculate each contribution and sum components using their signs.

Check dimensions:

G

has units

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

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

\mathrm{m/s^2}

. Check direction: attraction points toward the source. Check limiting behaviour: doubling the separation makes the force one quarter as large; doubling either mass doubles the force; and as separation grows without bound, the force approaches zero.

The gravitational law gives the gravitational force, not necessarily the net force if other forces act. To find an object’s actual acceleration, include all forces in the force sum before applying Newton’s second law. The gravitational acceleration itself remains the gravitational force per unit mass, directed toward the source.

\sum\mathbf F=m\mathbf a

Use centre-to-centre separation for the point-mass model.
Verify units and direction as well as the numerical value.
Keep the equal-and-opposite forces on the two masses distinct.

Worked example

Two particles on a line

Mass A is

12\ \mathrm{kg}

at

x=0

, and mass B is

5.0\ \mathrm{kg}

at

x=0.40\ \mathrm{m}

. Find the gravitational force on A. Use a ground-based Cartesian frame with the

x

-axis positive from A toward B. The velocities are not specified and are not needed to calculate this force.

Force on A

The attractive force on A points toward B.

Set up the system and separation
The system is particle A; B is the source. Their velocities do not affect the gravitational-force calculation. The centre-to-centre separation is the difference between their stated positions.
$r=0.40\ \mathrm{m}$
Apply the universal law
The law gives the magnitude. Since B lies in the positive $x$ direction from A, attraction makes the force component on A positive.
$F=G\frac{m_A m_B}{r^2}=6.6743\times10^{-11}\frac{(12)(5.0)}{(0.40)^2}\ \mathrm{N}$
Report and check
The force is approximately $2.50\times10^{-8}\ \mathrm{N}$ in the positive $x$ direction. The units reduce to newtons, and the direction is toward B.
$\mathbf F_A=+2.50\times10^{-8}\,\mathbf i\ \mathrm{N}$

Answer: The gravitational force on A is

2.50\times10^{-8}\ \mathrm{N}

toward B, in the positive

x

direction.

Check: The force on B has the same magnitude in the negative

x

direction. The two forces act on different particles.

Worked example

Force and gravitational acceleration near a spherical planet

A

2.0\ \mathrm{kg}

instrument is at a distance

6.80\times10^6\ \mathrm{m}

from the centre of a planet with mass

5.50\times10^{24}\ \mathrm{kg}

. Find the planet’s gravitational force on the instrument and the gravitational acceleration at its location. Take upward, away from the planet’s centre, as positive.

Instrument attracted to planet

The gravitational force points toward the planet’s centre.

Identify the system and source
The system is the instrument, treated as a particle; the planet is the source. The given distance is measured from the planet’s centre. The gravitational force points downward, opposite the positive direction.
Calculate the force magnitude
Substitute the masses and centre-to-centre distance into the universal law. The force component is negative because it points downward.
$F_g=G\frac{Mm}{r^2}=6.6743\times10^{-11}\frac{(5.50\times10^{24})(2.0)}{(6.80\times10^6)^2}=15.9\ \mathrm{N}$
Find gravitational acceleration
Gravitational acceleration is the planet’s gravitational force per unit mass, so its signed component is downward. This describes the gravitational field at the instrument’s location; it does not, by itself, determine the instrument’s actual acceleration if other forces also act.
$g_y=\frac{F_{g,y}}{m}=\frac{-15.9\ \mathrm{N}}{2.0\ \mathrm{kg}}=-7.94\ \mathrm{m/s^2}$

Answer: The planet’s gravitational force is

15.9\ \mathrm{N}

toward its centre. The gravitational acceleration at the instrument’s location is

7.94\ \mathrm{m/s^2}

toward the centre.

Check: The gravitational acceleration magnitude is also given by

GM/r^2

, independent of the instrument’s mass. Its units are

\mathrm{m/s^2}

, and the negative component agrees with the upward-positive axis. Other forces, if present, would affect the instrument’s actual acceleration, not the gravitational acceleration defined here.

Worked example

Adding forces from two sources

A

1.0\ \mathrm{kg}

particle is at the origin. A

4.0\ \mathrm{kg}

source mass is at

(0.30,0)\ \mathrm{m}

, and a

3.0\ \mathrm{kg}

source mass is at

(0,0.40)\ \mathrm{m}

. Find the net gravitational force on the particle. Use a fixed Cartesian frame with positive

x

right and positive

y

upward. The source velocities are not needed for this force calculation.

Two gravitational pulls

The net force is the vector sum of the two attractions.

Find each force
The system is the particle at the origin. The first source attracts it along positive $x$ ; the second attracts it along positive $y$ . The source coordinates give the distance to each mass.
$F_x=G\frac{(1.0)(4.0)}{(0.30)^2}=2.97\times10^{-9}\ \mathrm{N},\quad F_y=G\frac{(1.0)(3.0)}{(0.40)^2}=1.25\times10^{-9}\ \mathrm{N}$
Sum components
The forces are perpendicular, so add their components rather than their magnitudes. Both components are positive under the selected axes.
$\mathbf F_{\mathrm{net}}=(2.97\,\mathbf i+1.25\,\mathbf j)\times10^{-9}\ \mathrm{N}$
Find magnitude and direction
Use the component right triangle to find the net magnitude and the angle measured counterclockwise from positive $x$ .
$F_{\mathrm{net}}=3.22\times10^{-9}\ \mathrm{N},\qquad \theta=\tan^{-1}\!\left(\frac{1.25}{2.97}\right)=22.8^\circ$

Answer: The net force is

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

at

22.8^\circ

above the positive

x

-axis.

Check: The direction lies between the two source directions, as expected. The net magnitude is greater than either component and less than their arithmetic sum.

Common mistakes and how to avoid them

Using an object’s height above a planet’s surface as its distance from the planet’s centre.

Correction: For a spherical source treated as a point mass outside it, measure from the centre. Add the surface height to the planet’s radius when needed.

Treating the inverse-square law as an inverse-distance law.

Correction: The force varies as

1/r^2

. Doubling separation makes the force one quarter as large.

Adding force magnitudes from sources in different directions.

Correction: Resolve each force into components and add components with their signs.

Calling mass and weight the same thing.

Correction: Mass is measured in kilograms. Weight is a gravitational force measured in newtons and depends on the local gravitational acceleration.

Assuming gravitational acceleration is always the object’s actual acceleration.

Correction: Gravitational acceleration describes the gravitational force per unit mass. Actual acceleration follows from the net force, including any other forces acting.

Lesson summary

Two masses attract with magnitude $F=Gm_1m_2/r^2$ .
The force acts along the line joining the masses and points toward the source.
Use vectors and superposition when multiple source masses act.
For a spherical source outside its surface, gravitational acceleration has magnitude $GM/r^2$ and points toward its centre.
Check SI units, signs, direction, and inverse-square limiting behaviour.

Check your understanding

Question 1

Two point masses remain unchanged, but their separation is tripled. What happens to the gravitational-force magnitude?

It becomes one third as large.
It becomes one ninth as large.
It becomes three times as large.
It remains unchanged.

Show answer and explanation

It becomes one ninth as large.

The law contains the inverse square of separation, so tripling

r

divides the force by

3^2=9

.

Question 2

A source mass lies directly to the left of a particle. With positive

x

to the right, which way is the gravitational force on the particle?

Positive $x$ .
Negative $x$ .
Positive $y$ .
Its direction cannot be determined.

Show answer and explanation

Negative

x

.

Gravity is attractive, so the force points toward the source, which is in the negative

x

direction.

Key terms

Centre-to-centre separation: The straight-line distance between the centres of two masses in the point-mass model.
Gravitational constant: The universal constant $G$ that sets the strength of gravitational interaction.
Gravitational acceleration: Gravitational force per unit mass at a location; it points toward the source mass.
Weight: The gravitational force acting on an object.
Superposition: The rule that the total gravitational force is the vector sum of forces from each source mass.

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4.1 · Apply Newton’s law of universal gravitation

University of Alberta EN PH 131: Engineering Mechanics: Dynamics

EN PH 131 study topic 4.1

What you will learn

1. The gravitational-force model

2. Coordinates, signs, and gravitational acceleration

3. A reliable solution and verification process

Two particles on a line

Force and gravitational acceleration near a spherical planet

Adding forces from two sources

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