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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.
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 HammerFeather 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 = 
 = m g 
so 


g = 
 . 
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.
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 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 d^{2}/G. Close to the
Earth's surface at a distance of 6.4 × 10^{6} meters from the center,
g = 9.8 meters/second^{2}. The distance is given in meters to match the units of the
gravity accelerationwhen 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} m^{3}/(kg sec^{2}).
Plugging in the values, you will find the Earth's mass = 9.8 ×
(6.4×10^{6})^{2} / (6.7 × 10^{11}) kilograms =
6.0 × 10^{24} 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
 Gravitational Acceleration: g = (G × Mass)/(distance from the center)^{2}.
 Comparing gravitational accelerations: acceleration at position A = acceleration at
position B × (distance B/distance A)^{2}.
 Calculating mass: Mass = (g × distance^{2})/G.
 What did Galileo discover about how objects of different masses fall to the
Earth?
 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.
 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.
 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.?
 Why do astronauts in orbit around the Earth feel ``weightless'' even though
the Earth's gravity is still very much present?
 Put the following in order of their acceleration around the Earth: a 200ton
space station 6580 kilometers from the center, a 60kilogram astronaut
6580 kilometers
from the center, a 1ton satellite 418,000 kilometers from the center, and the
7.4×10^{19}ton Moon 384,000 kilometers from the center. Explain your answer.
 How can you find the mass of the Earth using ordinary objects in your house?
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last updated:
September 8, 2013
Is this page a copy of Strobel's
Astronomy Notes?
Author of original content:
Nick Strobel