Lesson Explainer: Newtonโ€™s Law of Universal Gravitation Mathematics

In this explainer, we will learn how to apply Newtonโ€™s law of universal gravitation to find the gravitational force between two masses.

Consider two masses ๐‘š๏Šง and ๐‘š๏Šจ separated by a distance ๐‘Ÿ. Then, each of these masses exerts a force of attraction on the other called a gravitational force. This force occurs between any pair of objects with nonzero masses and is inversely proportional to the square of the distance between the two masses. This force occurs as a direct result of Newtonโ€™s third law of motion.

Definition: Newtonโ€™s Third Law of Motion

Newtonโ€™s third law of motion states that when one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

For the gravitational forces exerted by a pair of bodies, these conditions can be supplemented by some additional conditions, which together constitute a definition of Newtonโ€™s law of universal gravitation.

Let us now define Newtonโ€™s law of universal gravitation.

Definition: Newtonโ€™s Law of Universal Gravitation

Two bodies exert gravitational forces on each other, where the direction of the force on either body is toward the center of mass of the other body.

The magnitudes of the forces are given by ๐น=๐บ๐‘š๐‘š๐‘Ÿ,๏Šง๏Šจ๏Šจ where ๐น is the force measured in newtons, ๐‘š๏Šง and ๐‘š๏Šจ are the masses of the bodies measured in kilograms, ๐‘Ÿ is the distance between the centers of mass of the bodies measured in metres, and ๐บ is a constant, where ๐บโ‰ˆ6.67ร—10โ‹…/.๏Šฑ๏Šง๏Šง๏Šจ๏ŠจNmkg

๐บ is known as the universal gravitational constant.

Note: If two objects must be in contact for a force to act, the force is known as a contact force. If this is not a necessity, the forces are called noncontact forces. Gravitational force is an example of a noncontact force. Sometimes, these noncontact forces act between particles that are in contact.

Letโ€™s look at an example in which the gravitational force generated between two bodies is determined.

Example 1: Calculating the Gravitational Force Between Two Bodies

Determine the gravitational force between two identical balls each of mass 3.01 kg, given that the distance between their centers is 15.05 cm and the universal gravitational constant is 6.67ร—10๏Šฑ๏Šง๏Šง Nโ‹…m2/kg2.

Answer

The force can be determined using the formula ๐น=๐บ๐‘š๐‘š๐‘Ÿ.๏Šง๏Šจ๏Šจ

The distance between the center of mass of the balls, ๐‘Ÿ, must be measured in metres to be consistent with the unit for the universal gravitational constant, ๐บ. The value of ๐‘Ÿ is then 0.1505 m.

Substituting the values given in the question, we find that ๐น=6.67ร—10๏€พ3.010.1505๏Š=2.668ร—10.๏Šฑ๏Šง๏Šง๏Šจ๏Šจ๏Šฑ๏ŠฎN

Let us now look at an example in which the gravitational force between two bodies is used to determine the distance between their centers of mass.

Example 2: Calculating the Distance Between Two Bodies Given the Gravitational Force Between them

Given that the gravitational force between two bodies of masses 4.6 kg and 2.9 kg was 3.2ร—10๏Šฑ๏Šง๏Šฆ N, find the distance between their centers. Take the universal gravitational constant ๐บ=6.67ร—10โ‹…/๏Šฑ๏Šง๏Šง๏Šจ๏ŠจNmkg.

Answer

The formula for determining the gravitational force between bodies, ๐น=๐บ๐‘š๐‘š๐‘Ÿ,๏Šง๏Šจ๏Šจ can be rearranged to make ๐‘Ÿ the subject as follows: ๐‘Ÿ=๐บ๐‘š๐‘š๐น๐‘Ÿ=๏„ž๐บ๐‘š๐‘š๐น.๏Šจ๏Šง๏Šจ๏Šง๏Šจ

Substituting the values given in the question, we find that ๐‘Ÿ=๏„ž6.67ร—104.6(2.9)3.2ร—10๐‘Ÿ=โˆš6.67ร—10(4.16875ร—10)๐‘Ÿ=โˆš2.78055625๐‘Ÿ=1.6675.๏Šฑ๏Šง๏Šง๏Šฑ๏Šง๏Šฆ๏Šฑ๏Šง๏Šง๏Šง๏Šฆm

The distance between the centers of the bodies is 166.75 cm.

Let us look now at an example in which the gravitational force generated between two bodies is determined, where one of the bodies is Earth.

Example 3: Calculating the Gravitational Force Between Earth and a Satellite

A satellite of mass 2โ€Žโ€‰โ€Ž415 kg is orbiting Earth 540 km above its surface. Given that the universal gravitational constant is 6.67ร—10๏Šฑ๏Šง๏Šง Nโ‹…m2/kg2 and Earthโ€™s mass and radius are 6ร—10๏Šจ๏Šช kg and 6โ€Žโ€‰โ€Ž360 km, determine the gravitational force exerted by Earth on the satellite.

Answer

The force can be determined using the formula ๐น=๐บ๐‘š๐‘š๐‘Ÿ.๏Šง๏Šจ๏Šจ

The distance between Earth and the satellite, ๐‘Ÿ, must be measured in metres to be consistent with the value of the universal gravitational constant, ๐บ.

Since ๐‘Ÿ in the formula is the distance between the centers of mass, the value of ๐‘Ÿ is the sum of the distance from the satellite to the surface of Earth and the radius of Earth: ๐‘Ÿ=(6360+540)ร—10=6.9ร—10.๏Šฉ๏Šฌ

Substituting the values given in the question, we find that ๐น=6.67ร—10๏€ฟ2415๏€น6ร—10๏…(6.9ร—10)๏‹๐น=6.67ร—10๏€ฟ1.449ร—10(6.9ร—10)๏‹๐น=6.67ร—10๏€น3.0434783ร—10๏…=20300.๏Šฑ๏Šง๏Šง๏Šจ๏Šช๏Šฌ๏Šจ๏Šฑ๏Šง๏Šง๏Šจ๏Šฎ๏Šฌ๏Šจ๏Šฑ๏Šง๏Šง๏Šง๏ŠชN

The acceleration of a body due to the gravitational force between itself and another body can be determined by equating the force between the bodies and the net force on the body whose acceleration is being determined. We have then that ๐น=๐บ๐‘š๐‘š๐‘Ÿ,๏Šง๏Šจ๏Šจ and ๐น=๐‘š๐‘Ž,๏Šง where the body whose acceleration is determined has mass ๐‘š๏Šง . From this we obtain ๐‘š๐‘Ž=๐บ๐‘š๐‘š๐‘Ÿ.๏Šง๏Šง๏Šจ๏Šจ

Since ๐‘šโ‰ 0๏Šง, we can divide through by ๐‘š๏Šง. Hence, ๐‘Ž=๐บ๐‘š๐‘Ÿ.๏Šจ๏Šจ

We can see that the acceleration of the body of mass ๐‘š๏Šง does not depend on the value of ๐‘š๏Šง.

Definition: The Gravitational Field Strength of a Point Mass

Gravitational field strength is the gravitational force per unit mass exerted by a mass, ๐‘š, on a body whose center of mass is a distance ๐‘Ÿ from ๐‘š. It is given by ๐‘”=๐บ๐‘š๐‘Ÿ.๏Šจ

Gravitational field strength is a property of any point in a field, whether in contact with the surface of the gravitating body or not.

On Earth, the acceleration due to gravity is approximately 9.8 m/s2. This value, known as ๐‘”, can be determined from the mass, ๐‘š=5.97ร—10๏Šจ๏Šชkg, and radius, ๐‘Ÿ=6.37ร—10๏Šฌm, of Earth as follows: ๐‘”=6.67ร—10๏€ฟ5.97ร—10(6.37ร—10)๏‹,๏Šฑ๏Šง๏Šง๏Šจ๏Šช๏Šฌ๏Šจ which is 9.8 to one decimal place.

Let us now look at an example in which the ratio of the acceleration due to gravity on Earth and another planet is determined.

Example 4: Comparing the Acceleration Due to Gravity on Earth and Another Planet

Given that a planetโ€™s mass and diameter are 3 and 6 times those of Earth, respectively, calculate the ratio between the acceleration due to gravity on that planet and that on Earth.

Answer

The acceleration due to gravity at the surface of a uniform sphere can be determined using the formula ๐‘Ž=๐บ๐‘š๐‘Ÿ,๏Šจ where ๐‘š is the mass of the sphere and ๐‘Ÿ is its radius.

Taking ๐‘š as the mass of Earth and ๐‘Ÿ as the radius of Earth, the acceleration due to gravity on a planet with a mass 3 times that of Earth and a diameter 6 times that of Earth is expressed by ๐‘Ž=๐บ3๐‘š(6๐‘Ÿ),๏Šง๏Šจ since multiplying the diameter of a sphere by a constant multiplies the radius of the sphere by the same constant. The expression for acceleration due to gravity at the planet can be expressed as the acceleration due to gravity on Earth multiplied by a constant, as follows: ๐‘Ž=๐บ3๐‘š36๐‘Ÿ=๐บ๐‘š12๐‘Ÿ=112๐บ๐‘š๐‘Ÿ.๏Šง๏Šจ๏Šจ๏Šจ

Since the acceleration due to gravity on the planet is 112 of that on Earth, the ratio of the acceleration due to gravity on the planet to the acceleration due to gravity on Earth is 1โˆถ12.

Let us now look at an example in which the radius of a planet is determined from the acceleration due to gravity on the planet.

Example 5: Finding the Radius of a Planet Given Information about a Dropped Object

An astronaut dropped an object from a height of 2โ€Žโ€‰โ€Ž352 cm above the surface of a planet, and it reached the surface after 8 s. The mass of the planet is 7.164ร—10๏Šจ๏Šช kg, while that of Earth is 5.97ร—10๏Šจ๏Šช kg, and the radius of Earth is 6.34ร—10๏Šฌ m. Given that the gravitational acceleration of Earth is ๐‘”=9.8/ms๏Šจ, find the radius of the other planet.

Answer

The acceleration near the surface of the planet can be approximated using the acceleration due to gravity of the planet on its surface. Then, the acceleration due to gravity at the surface of a uniform sphere can be determined using the formula ๐‘Ž=๐บ๐‘š๐‘Ÿ,๏Šจ where ๐‘š is the mass of the sphere and ๐‘Ÿ is its radius.

The acceleration due to gravity on the planet the astronaut visits can be determined from the motion of the dropped object using the formula ๐‘ =๐‘ข๐‘ก+12๐‘Ž๐‘ก,๏Šจ where ๐‘  is the displacement of the dropped object, ๐‘ข is the initial velocity of the object, ๐‘ก is the time that the object moves for, and ๐‘Ž is the acceleration of the object.

The object is dropped from a point 2โ€Žโ€‰โ€Ž352 cm above the surface of the planet. To be consistent with the universal gravitational constant, this displacement is converted to a value in metres of 23.52 m. The object is dropped from rest; hence, we have that 23.52=12๐‘Ž๏€น8๏…,๐‘Ž=2(23.52)64=0.735/.๏Šจ๏Šจms

We can make ๐‘Ÿ the subject of the formula, ๐‘Ž=๐บ๐‘š๐‘Ÿ,๏Šจ giving us ๐‘Ÿ=๐บ๐‘š๐‘Ž,๐‘Ÿ=๏„ž๐บ๐‘š๐‘Ž.๏Šจ

Substituting the acceleration determined, the known mass of the planet, and taking ๐บ=6.67ร—10โ‹…/,๏Šฑ๏Šง๏Šง๏ŠจNmkg we have ๐‘Ÿ=๏„Ÿ6.67ร—10๏€ฝ7.164ร—100.735๏‰,๐‘Ÿโ‰ˆ2.55ร—10.๏Šฑ๏Šง๏Šง๏Šจ๏Šช๏Šญm

It is important to note that the question does not provide a value of ๐บ to use.

A method of determining the radius of the planet without knowledge of the value of ๐บ can be obtained by comparing the expressions for the accelerations due to gravity on the planet and on Earth.

On Earth we have that ๐‘Ž=๐บ๐‘š๐‘Ÿ,EEE๏Šจ where ๐‘ŽE is the acceleration due to gravity on Earth, ๐‘šE is the mass of Earth, and ๐‘ŸE is the radius of Earth.

Equivalently, on the planet we have that ๐‘Ž=๐บ๐‘š๐‘Ÿ,PPP๏Šจ where ๐‘ŽP is the acceleration due to gravity on the planet, ๐‘šP is the mass of the planet, and ๐‘ŸP is the radius of the planet.

We can divide ๐‘ŽE by ๐‘ŽP to obtain ๐‘Ž๐‘Ž=๐บ๏€ผ๏ˆ๐บ๏€ผ๏ˆ๐‘Ž๐‘Ž=๏€ผ๏ˆ๏€ผ๏ˆ๐‘Ž๐‘Ž=๏€ฝ๐‘š๐‘š๏‰๏€พ๐‘Ÿ๐‘Ÿ๏Š.EPEPEPEPPE๏‰๏Ž๏‰๏Ž๏‰๏Ž๏‰๏Ž๏Šจ๏ŠจEEPPEEPP๏Žก๏Žก๏Žก๏Žก

We can make ๐‘ŸP the subject of the expression, as follows: ๐‘Ÿ=๏€ฝ๐‘Ž๐‘Ž๏‰๏€ฝ๐‘š๐‘š๏‰๐‘Ÿ๐‘Ÿ=๏„Ÿ๏€ฝ๐‘Ž๐‘Ž๏‰๏€ฝ๐‘š๐‘š๏‰๐‘Ÿ.PEPPEEPEPPEE๏Šจ๏Šจ๏Šจ

Substituting the values of ๐‘ŽE, ๐‘šE, ๐‘ŸE, ๐‘šP, and ๐‘ŽP, we obtain ๐‘Ÿ=๏„Ÿ๏€ผ9.80.735๏ˆ๏€ฝ7.164ร—105.97ร—10๏‰(6.34ร—10)๐‘Ÿ=๏„ž๏€ผ9.80.735๏ˆ๏€ผ7.1645.97๏ˆ(6.34ร—10)๐‘Ÿ=2.536ร—10.PPP๏Šจ๏Šช๏Šจ๏Šช๏Šฌ๏Šจ๏Šฌ๏Šจ๏Šญm

This value of ๐‘ŸP is slightly different to that obtained using the standard value of ๐บ, as the value calculated comparing the expressions for acceleration due to gravity on Earth and on the planet depends on the precision used in stating the value of the radius and mass of Earth rather than on the value of the universal gravitational constant.

It is important to note that the kinematic equation used to determine ๐‘ŽP is only approximately correct, as the acceleration due to gravity at a point 23.52 m above the surface of the planet does not equal the acceleration due to gravity at the surface of the planet. Hence, the value for acceleration due to gravity on the planet determined by dropping an object from above the surface of the planet will underestimate the acceleration and hence overestimate the radius of the planet.

Let us now summarize some of the key concepts from this explainer.

Key Points

  • Two bodies exert gravitational forces on each other, where the direction of the force on either body is toward the center of mass of the other body. The magnitudes of the forces are given by ๐น=๐บ๐‘š๐‘š๐‘Ÿ,๏Šง๏Šจ๏Šจ where ๐น is the force measured in newtons, ๐‘š๏Šง and ๐‘š๏Šจ are the masses of the bodies measured in kilograms, ๐‘Ÿ is the distance between the centers of mass of the bodies measured in metres, and ๐บ is a constant, where ๐บโ‰ˆ6.67ร—10โ‹…/.๏Šฑ๏Šง๏Šง๏ŠจNmkg๐บ is known as the universal gravitational constant.
  • The acceleration of a body due to the gravitational force between itself and another body can be determined by equating the force between the bodies and the net force on the body whose acceleration is being determined: ๐‘š๐‘Ž=๐บ๐‘š๐‘š๐‘Ÿ.๏Šง๏Šง๏Šจ๏Šจ Hence, ๐‘Ž=๐บ๐‘š๐‘Ÿ,๏Šจ๏Šจ where ๐‘Ž is the acceleration due to gravity of a body of mass ๐‘š๏Šจ and ๐‘Ÿ is the distance between the centers of mass of ๐‘š๏Šง and๐‘š๏Šจ.
  • For two bodies that have masses ๐‘š๏Šง and ๐‘š๏Šจ and radii ๐‘Ÿ๏Šง and ๐‘Ÿ๏Šจ, the ratio of the accelerations due to gravity of these bodies is given by ๐‘Ž๐‘Ž=๏€ฝ๐‘š๐‘š๏‰๏€พ๐‘Ÿ๐‘Ÿ๏Š.๏Šง๏Šจ๏Šง๏Šจ๏Šจ๏Šจ๏Šง๏Šจ

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