Gravitational Acceleration

Chapter index in this window —   — Chapter index in separate window

Video lecture for this chapter

This material (including images) is copyrighted!. See my copyright notice for fair use practices.

Galileo found that the acceleration due to gravity (called ``g'') depends only on the mass of the gravitating object and the distance from it. It does not depend on the mass of the object being pulled. In the absence of air drag, a huge boulder will fall at the same rate as a small marble dropped from the same height as the boulder. A tiny satellite at the same distance from the Sun as Jupiter's orbit from the Sun feels the same acceleration from the Sun as the large planet Jupiter does from the Sun. (The satellite and Jupiter will also, therefore, have the same orbital period around the Sun.) How is this possible? Most people would agree with Aristotle that the bigger object should fall faster than the smaller object, but experiments show they would be wrong. One particularly nice illustration of this was done on the Moon by Commander David Scott of Apollo 15 when he dropped a hammer and a feather from the same height. See the video on the Apollo 15 Hammer-Feather Drop page.

A boulder falling toward the Earth is pulled by a stronger gravity force than the marble, since the boulder's mass is greater than the marble, but the boulder also has greater resistance to a change in its motion because of its larger mass. The effects cancel each other out, so the boulder accelerates at the same rate as the marble. The same line of reasoning explains the equal acceleration experienced by Jupiter and the satellite.

You can use Newton's second law of motion F = m × a (which relates the acceleration, a, felt by a object with mass m when acted on by a force F) to derive the acceleration due to gravity (here replace a with g) from a massive object:

The force of gravity =
(G M m)
d2
= m g
so

g =
(G M)
d2
.
The gravitational acceleration depends on only the mass of the gravitating object M and the distance d from it. Notice that the mass of the falling object m has been cancelled out. This explains why astronauts orbiting the Earth feel ``weightless''. In orbit they are continually ``falling'' toward the Earth because of gravity (the Earth's surface curves away from them at the same rate they are moving forward). If Jane Astronaut drops a pen in the space shuttle, it accelerates toward the Earth, but she accelerates by the same amount so the pen remains at the same position relative to her. In fact the entire shuttle and its contents are accelerating toward the Earth at the same rate, so Jane and her companions ``float'' around inside! This is because all of them are at very nearly the same distance from the Earth.

same acceleration produces weightlessness

The acceleration decreases with the SQUARE of the distance (inverse square law). To compare gravity accelerations due to the same object at different distances, you use the gravity acceleration g at distance A = (the gravity acceleration g at distance B) × (distance B / distance A)2. Notice which distance is in the top of the fraction. An example of using the inverse square law is given in the ``How do you do that?'' box below.

How do you do that?

Find how many times more gravitational acceleration the Galileo atmosphere probe felt at 100,000 miles from Jupiter's center than the orbiter felt at 300,000 miles. You have
probe's g = orbiter's g × (300,000/100,000)2 = orbiter's g × (3/1)2

= orbiter's g × 9.
The probe accelerated by an amount nine times greater than the orbiter.

Measuring the Mass of the Earth

Measuring the acceleration of an object dropped to the ground enables you to find the mass of the Earth. You can rearrange the gravity acceleration relation to solve for the mass M to find M = g d2/G. Close to the Earth's surface at a distance of 6.4 × 106 meters from the center, g = 9.8 meters/second2. The distance is given in meters to match the units of the gravity acceleration---when you do a calculation, you must be sure you check that your units match up or you will get nonsense answers. The big G is the universal gravitational constant, approximately 6.7×10-11 m3/(kg sec2). Plugging in the values, you will find the Earth's mass = 9.8 × (6.4×106)2 / (6.7 × 10-11) kilograms = 6.0 × 1024 kilograms. If you are unsure of how to work with scientific notation, read the scientific notation section in the mathematics review appendix (pay close attention to the part describing how to enter scientific notation on your calculator!).

You can determine masses of stars and planets in a similar way: by measuring the acceleration of objects orbiting them and the distance between the star or planet and the object. A small object falling to the Earth has mass and, therefore, has a gravitational acceleration associated with it: the Earth is accelerated toward the falling object (an example of Newton's third law)! However, if you plug some typical masses of terrestrial objects (less than, say, 1000 kilograms) into the acceleration formula, you will see that the amount the Earth is accelerated is vastly smaller than the falling object's acceleration. You can ignore the Earth's acceleration.

A side note: determining the mass of the Earth also depends on knowing the value of the gravitational constant G. The constant was first measured by Henry Cavendish in 1798. After discussing his experimental results, he then applied his measurement to the subject of his paper's title: ``Weighing the Earth.''

Formulae

  1. Gravitational Acceleration: g = (G × Mass)/(distance from the center)2.
  2. Comparing gravitational accelerations: acceleration at position A = acceleration at position B × (distance B/distance A)2.
  3. Calculating mass: Mass = (g × distance2)/G.

Review Questions

  1. What did Galileo discover about how objects of different masses fall to the Earth?
  2. If you dropped a hammer and feather from the same height above the Earth's surface, which would actually hit the ground first? Why would it be different than what Galileo said about falling objects? Explain why if you let the feather fall quill end first, the result is closer to what Galileo said.
  3. If you dropped a hammer and feather from the same height above the airless Moon's surface, which would actually hit the ground first? Explain why your answer is different than for the previous question.
  4. How many times less/more gravity acceleration due to the Sun does the Ulysses spacecraft feel at 2.3 A.U. above the Sun than the solar gravity acceleration it felt at Jupiter (5.2 A.U. from the Sun)? Is it accelerated more or less at 2.3 A.U. than when it was at 5.2 A.U.?
  5. Why do astronauts in orbit around the Earth feel ``weightless'' even though the Earth's gravity is still very much present?
  6. Put the following in order of their acceleration around the Earth: a 200-ton space station 6580 kilometers from the center, a 60-kilogram astronaut 6580 kilometers from the center, a 1-ton satellite 418,000 kilometers from the center, and the 7.4×1019-ton Moon 384,000 kilometers from the center. Explain your answer.
  7. How can you find the mass of the Earth using ordinary objects in your house?

previousGo back to previous section -- next Go to next section

Go to Astronomy Notes home

last updated: February 21, 2022

Is this page a copy of Strobel's Astronomy Notes?

Author of original content: Nick Strobel