D2.2 Solve universal-gravitation and circular-orbit problems

Ontario Grade 12 Physics

Gravitational, Electric, and Magnetic Fields

SPH4U expectation D2.2

In SPH3U, you used force to explain changes in motion and treated velocity as a vector with both magnitude and direction. This lesson applies those ideas to gravity and circular motion. Before calculating, identify the physical system, the reference frame, and the positive direction. A reference frame is the viewpoint used to describe motion. For a satellite orbiting Earth, use a frame centred on Earth and treat the satellite as the moving object. Gravity acts toward Earth’s centre. For circular-motion calculations, take inward, toward the centre, as the positive radial direction. Force and velocity are vectors; mass, distance, speed, and period are scalars.

What you will learn

Use Newton’s law of universal gravitation to calculate the force between two masses.
Determine how gravity provides the inward force needed for a circular orbit.
Solve for orbital speed or period using a satellite’s distance from the centre of its orbiting body.
Check units, direction, significant figures, and whether an answer is reasonable.

1. Universal gravitation: the force between two masses

Newton’s law of universal gravitation describes the attractive force between any two objects with mass. In a simple model, represent each object as a point mass. The separation

r

is the distance between their centres, not the distance between their surfaces. The gravitational force is directed along the line joining the two centres, and each object pulls on the other.

The force magnitude increases if either mass increases. It decreases as the square of the centre-to-centre distance increases. This inverse-square relationship means that doubling the distance makes the force one-quarter as large. The gravitational constant is

G=6.67\times10^{-11}\ \mathrm{N\,m^2/kg^2}

. Use masses in kilograms and distance in metres.

A force is a vector, so state its direction as well as its magnitude. In a two-object problem, the forces on the two objects have equal magnitudes and opposite directions. Do not confuse the force between two objects with the gravitational force on an object near a planet; both are found using the same law.

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

Use centre-to-centre distance for $r$ .
The gravitational force is attractive and points toward the other mass.
Keep mass in kilograms and distance in metres.

2. Circular motion and gravity as the inward force

An object moving in a circle has changing velocity direction, even if its speed stays constant. Its acceleration points toward the centre of the circle. The inward acceleration is called centripetal acceleration; “centripetal” means centre-seeking. For speed

v

and circular-path radius

r

, its magnitude is

v^2/r

.

A free-body diagram for a satellite in an ideal circular orbit has one force: Earth’s gravitational force, directed inward. There is no separate outward force in this model. The inward net force required for circular motion is called centripetal force; it is the name for the net inward force, not an extra force to add to the diagram.

For a satellite of mass

m

orbiting a body of mass

M

, set the gravitational force equal to the required inward net force. The satellite’s mass appears on both sides and cancels, so the orbital speed depends on the central body’s mass and the orbit radius. Here,

r

is measured from the centre of the central body. If the orbit is described by altitude above a surface, first add the body’s radius.

The period

T

is the time for one complete orbit. One orbit covers a circumference of

2\pi r

. Dividing this distance by the constant orbital speed gives the period. Use seconds for the SI unit of time; convert to minutes only after calculating.

G\frac{Mm}{r^2}=m\frac{v^2}{r},\qquad v=\sqrt{\frac{GM}{r}},\qquad T=\frac{2\pi r}{v}

Choose inward as positive for the radial force equation.
For a circular orbit, gravity supplies the net inward force.
An orbit radius is measured from the centre, not the surface.

3. A reliable problem-solving approach

Start by naming the objects and choosing a reference frame. For a planet–satellite system, state which mass is the central body and which object is orbiting. Identify the known values and the requested quantity. Sketch the centres, the separation, and any direction needed. For an orbit, draw the satellite on a circle and show an inward arrow labelled gravitational force.

Next, select the relationship that matches the question. Use universal gravitation for the force between masses. For a circular orbit, equate gravitational force and the inward net force. Rearrange algebraically before substituting values. This makes it easier to see which quantities are required and helps prevent mixing up radius and altitude.

Substitute values with units, then calculate. Keep extra digits during intermediate steps and round the final result to a sensible number of significant figures based on the given data. A final check should include units, direction, and reasonableness. For example, an inward force must point toward the orbit’s centre; an orbital speed should be positive; and the force should become smaller when the separation is made larger.

r=R+h

Draw the force direction before using a sign convention.
Show units in the substitution and final answer.
Check whether changing a mass or distance should raise or lower the result.

Worked example

1. Force between two objects

Two objects have masses of

1.20\times10^3\ \mathrm{kg}

and

8.00\times10^2\ \mathrm{kg}

. Their centres are

4.00\ \mathrm{m}

apart. Find the gravitational force magnitude and state its direction.

Set the system and direction
The system is the two objects, treated as point masses. Use a straight-line frame along the line joining their centres. The force on either object points toward the other; choose that direction as positive when describing the force on the object of interest.
Choose the law
The known quantities are both masses and their centre-to-centre separation. The requested quantity is the force magnitude, so use universal gravitation.
$F_g=G\frac{m_1m_2}{r^2}$
Substitute and calculate
Use $G=6.67\times10^{-11}\ \mathrm{N\,m^2/kg^2}$ . The units reduce to newtons.
$F_g=(6.67\times10^{-11})\frac{(1.20\times10^3)(8.00\times10^2)}{(4.00)^2}=4.00\times10^{-6}\ \mathrm{N}$

Answer: The force magnitude is

4.00\times10^{-6}\ \mathrm{N}

. Each object’s force is directed toward the other object.

Check: The result has units of newtons. It is attractive, as required. A very small force is reasonable for objects with these masses separated by several metres.

Worked example

2. Gravitational force on a satellite

A

1.20\times10^3\ \mathrm{kg}

satellite is at a distance of

7.00\times10^6\ \mathrm{m}

from Earth’s centre. Use Earth’s mass,

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

, to find the gravitational force on the satellite.

Set the system and direction
The system is Earth and the satellite. Use an Earth-centred reference frame and take inward, toward Earth’s centre, as positive. The given distance is already from the centre, so it is the radius to use.
Choose the law
The known values are Earth’s mass, the satellite’s mass, and their separation. The force on the satellite is directed toward Earth.
$F_g=G\frac{Mm}{r^2}$
Substitute and calculate
Use SI units throughout. The calculated magnitude is rounded to three significant figures.
$F_g=(6.67\times10^{-11})\frac{(5.97\times10^{24})(1.20\times10^3)}{(7.00\times10^6)^2}=9.75\times10^3\ \mathrm{N}$

Answer: The gravitational force on the satellite is

9.75\times10^3\ \mathrm{N}

, directed inward toward Earth’s centre.

Check: The units reduce to newtons, and the direction is inward. The result is a force on a massive satellite near Earth, so a value of several thousand newtons is reasonable.

Worked example

3. Speed and period of a circular orbit

A satellite follows a circular orbit

4.00\times10^5\ \mathrm{m}

above Earth’s surface. Find its orbital speed and period. Use Earth’s radius,

6.37\times10^6\ \mathrm{m}

, and mass,

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

.

Set the system and radius
The system is Earth and the satellite in an Earth-centred frame. The inward direction is toward Earth’s centre. The orbit radius is the distance from that centre, so add the altitude to Earth’s radius.
$r=6.37\times10^6\ \mathrm{m}+4.00\times10^5\ \mathrm{m}=6.77\times10^6\ \mathrm{m}$
Find the orbital speed
For a circular orbit, gravity provides the inward net force. The satellite’s mass cancels, leaving a speed determined by Earth’s mass and the orbit radius.
$v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\times10^{-11})(5.97\times10^{24})}{6.77\times10^6}}=7.67\times10^3\ \mathrm{m/s}$
Find the period
One orbit covers a circumference of $2\pi r$ . Divide by the orbital speed, then convert the result from seconds to minutes.
$T=\frac{2\pi r}{v}=\frac{2\pi(6.77\times10^6\ \mathrm{m})}{7.67\times10^3\ \mathrm{m/s}}=5.55\times10^3\ \mathrm{s}=92.5\ \mathrm{min}$

Answer: The orbital speed is

7.67\times10^3\ \mathrm{m/s}

, and the period is

5.55\times10^3\ \mathrm{s}

, or

92.5\ \mathrm{min}

.

Check: The speed has units of metres per second and the period has units of time. Both are positive. A low Earth orbit has a period measured in roughly an hour and a half, so the result is reasonable.

Common mistakes and how to avoid them

Using altitude above a planet’s surface as the orbit radius.

Correction: Add the planet’s radius to the altitude. The orbit radius runs from the planet’s centre to the satellite.

Adding a separate outward force to a circular-orbit free-body diagram.

Correction: Show the actual force or forces. In the ideal orbit model, gravity is the inward net force; “centripetal force” describes that net inward role.

Treating the gravitational force as a directionless scalar.

Correction: The law gives a force magnitude. State that its direction is along the line between the masses and points toward the other mass.

Forgetting to convert given quantities into SI units.

Correction: Use kilograms, metres, seconds, and newtons in the equations. Convert units before substitution.

Lesson summary

Universal gravitation gives the attractive force between two masses using their masses and centre-to-centre separation.
In a circular orbit, gravity supplies the net inward force. The satellite’s mass cancels when finding orbital speed.
Use the centre-to-centre orbit radius, and calculate period from orbit circumference divided by speed.
Include units and direction, round sensibly, and check whether the result matches the physical situation.

Check your understanding

Question 1

If the centre-to-centre distance between two masses doubles while the masses stay constant, what happens to the gravitational force magnitude?

It becomes twice as large.
It becomes half as large.
It becomes one-quarter as large.
It stays the same.

Show answer and explanation

It becomes one-quarter as large.

The force is inversely proportional to the square of the distance. Doubling the distance multiplies the denominator by four, so the force becomes one-quarter as large.

Question 2

A satellite is in a circular orbit around a planet. In the ideal model, which way does the net force point?

Along the satellite’s direction of motion.
Away from the planet’s centre.
Toward the planet’s centre.
There is no net force because the speed is constant.

Show answer and explanation

Toward the planet’s centre.

The satellite’s velocity direction changes as it moves around the circle, so it has inward acceleration. Gravity supplies the net force toward the planet’s centre.

Question 3

A satellite’s altitude above Earth is

3.00\times10^5\ \mathrm{m}

. Earth’s radius is

6.37\times10^6\ \mathrm{m}

. What radius should be used in the circular-orbit equations?

$3.00\times10^5\ \mathrm{m}$
$6.07\times10^6\ \mathrm{m}$
$6.67\times10^6\ \mathrm{m}$
$6.37\times10^6\ \mathrm{m}$

Show answer and explanation

6.67\times10^6\ \mathrm{m}

The orbit radius is measured from Earth’s centre. Add the altitude to Earth’s radius to obtain

6.67\times10^6\ \mathrm{m}

.

Key terms

Reference frame: The viewpoint and coordinate directions used to describe position and motion.
Centre-to-centre distance: The separation measured from the centre of one object to the centre of the other.
Centripetal acceleration: The inward acceleration of an object moving along a circular path.
Centripetal force: The name for the net inward force that produces circular motion; it is not an additional force.
Orbital period: The time taken for one complete orbit.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation D2.2. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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D2.2 · Solve universal-gravitation and circular-orbit problems

Ontario Grade 12 Physics

SPH4U expectation D2.2

What you will learn

1. Universal gravitation: the force between two masses

2. Circular motion and gravity as the inward force

3. A reliable problem-solving approach

1. Force between two objects

2. Gravitational force on a satellite

3. Speed and period of a circular orbit

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH4U

About this lesson and its review