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AP®PhysicsCollege BoardPhysics 1: Algebra-BasedStudy GuidesUnit 2: Force & Translational DynamicsGravitational ForceNewton's Law of Gravitation

Newton's Law of Gravitation (College Board AP® Physics 1: Algebra-Based)

Study Guide

Ann Howell

Written by: Ann Howell

Reviewed by: Caroline Carroll

Newton's law of gravitation equation

  • Newton’s law of universal gravitation describes the gravitational force between two objects or systems as:

    • directly proportional to each of their masses

    • inversely proportional to the square of the distance between the systems’ centers of mass

  • Newton's law of universal gravitation applies between two bodies when outside a uniform gravitational field

    • For example, between the Earth and the Sun or the Earth and the Moon

  • Although planets are not point masses, their separation is much larger than their radius

    • Therefore, Newton’s law of gravitation applies to planets orbiting the Sun

  • Recall that the mass of a uniform sphere can be considered to be a point mass at its centre

    • So the gravitational field around a uniform sphere is identical to that around a point mass

  • The equation for Newton's law of universal gravitation is given by:

    open vertical bar stack F subscript g with rightwards arrow on top close vertical bar space equals space G fraction numerator m subscript 1 m subscript 2 over denominator r squared end fraction

  • Where:

    • open vertical bar stack F subscript g with rightwards arrow on top close vertical bar space equals magnitude of the gravitational force between the two objects, measured in straight N

    • G space equals spaceuniversal gravitational constant = 6.67 space cross times space 10 to the power of negative 11 end exponent space straight N times straight m squared space divided by space kg squared

    • m subscript 1 space equals spacemass of object 1, measured in kg

    • m subscript 2 space equals spacemass of object 2, measured in kg

    • r space equals spacethe distance between the centres of mass of the two objects, measured in straight m

Quantities in Newton's law of universal gravitation

Diagram showing gravitational force (Fg) between two masses, m1 and m2, at distance r. Arrows indicate the forces toward each other.
Newton's law of universal gravitation says that the gravitational force Fg is equal to G times the mass of both objects m1 and m2 divided by the square of the distance between the centres of mass r

  • The F subscript g space proportional to space 1 over r squaredrelation is called the ‘inverse square law’

    • This means that when the distance between masses is doubled, the gravitational force between the objects reduces by open parentheses 1 half close parentheses squared space equals space 1 fourth

Worked Example

A satellite of mass 6500 kg is orbiting the Earth at 2000 km above the Earth's surface. The gravitational force between them is 37 kN.

Which of the following is the mass of the Earth? The radius of the Earth is 6400 km.

A      6.02 space cross times space 10 to the power of 18 space kg

B      3.41 space cross times space 10 to the power of 23 space kg

C      3.50 space cross times space 10 to the power of 24 space kg

D      6.02 space cross times space 10 to the power of 24 space kg

The correct answer is D

Answer:

Step 1: Analyze the scenario

  • Draw a diagram to show all the known quantities

Gravitational forces between Earth and a satellite. Earth’s radius is 6400 km, the distance to the satellite is 2000 km, and m2 equals 6500 kg. The mass of the Earth m1 is unknown.

  • The object model is used in this scenario

  • r is the distance between the center of the Earth and the satellite equal to the radius of the Earth R plus the height of orbit of the satellite h

    • r space equals space R space plus space h

Step 2: List the known quantities

  • Mass of satellite, m subscript 2 space equals space 6500 space kg

  • Height of satellite orbit above the Earths surface, h space equals space 2000 space km space equals space 2 space cross times space 10 to the power of 6 space straight m

  • Magnitude of gravitational force between Earth and satellite, open vertical bar F subscript g close vertical bar space equals space 37 space kN space equals space 37 space cross times space 10 cubed space straight N

  • Radius of the Earth, R space equals space 6400 space km space equals space 6.4 space cross times space 10 to the power of 6 space straight m

  • Universal gravitational constant, G space equals space 6.67 space cross times space 10 to the power of negative 11 end exponent space straight m cubed space divided by space open parentheses kg space times space straight s squared close parentheses

Step 3: Calculate the distance between the center of the Earth and the satellite r

r space equals space R space plus space h

r space equals space open parentheses 6.4 space cross times space 10 to the power of 6 close parentheses space plus space open parentheses 2 space cross times space 10 to the power of 6 close parentheses

r space equals space 8.4 space cross times space 10 to the power of 6 space straight m

Step 4: Rearrange the equation for Newton's law of universal gravitation to make the mass of the Earth m subscript 1 the subject

open vertical bar stack F subscript g with rightwards arrow on top close vertical bar space equals space G fraction numerator m subscript 1 m subscript 2 over denominator r squared end fraction

open vertical bar stack F subscript g with rightwards arrow on top close vertical bar space times space r squared space equals space G m subscript 1 m subscript 2

fraction numerator open vertical bar stack F subscript g with rightwards arrow on top close vertical bar space times space r squared over denominator G m subscript 2 end fraction space equals space m subscript 1

Step 5: Substitute the known quantities into the rearranged form of the equation to calculate the mass of the Earth m subscript 1

m subscript 1 space equals space fraction numerator open parentheses 37 space cross times space 10 cubed close parentheses space open parentheses 8.4 space cross times space 10 to the power of 6 close parentheses squared over denominator open parentheses 6.67 space cross times space 10 to the power of negative 11 end exponent close parentheses space times space 6500 end fraction

m subscript 1 space equals space 6.02 space cross times space 10 to the power of 24 space kg

  • The answer is therefore D

Examiner Tips and Tricks

A common mistake in exams is to forget to add together the distance from the surface of the planet h and its radius R to obtain the value of r. The distance r is measured from the center of the mass, which is from the center of the planet.

Earth's radius (R) with the center of mass, the height of orbit (h) above Earth’s surface, and the orbital radius (r), calculated as R plus h.

Make sure to square the separation r in the equation!

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    Author
    Reviewer
    Ann Howell

    Author: Ann Howell

    Expertise: Physics Content Creator

    Ann obtained her Maths and Physics degree from the University of Bath before completing her PGCE in Science and Maths teaching. She spent ten years teaching Maths and Physics to wonderful students from all around the world whilst living in China, Ethiopia and Nepal. Now based in beautiful Devon she is thrilled to be creating awesome Physics resources to make Physics more accessible and understandable for all students, no matter their schooling or background.

    Caroline Carroll

    Author: Caroline Carroll

    Expertise: Physics Subject Lead

    Caroline graduated from the University of Nottingham with a degree in Chemistry and Molecular Physics. She spent several years working as an Industrial Chemist in the automotive industry before retraining to teach. Caroline has over 12 years of experience teaching GCSE and A-level chemistry and physics. She is passionate about creating high-quality resources to help students achieve their full potential.