According to legend, Galileo Galilei dropped two balls of different mass from the top of the leaning tower of Pisa in 1589. Whether or not this public experiment ever took place, Galileo was able to demonstrate that, contrary to Aristotle’s teaching, all bodies fall at the same rate regardless of mass, assuming that one is not so tenuous that it is slowed by air resistance. In this experiment, an equation is presented relating the acceleration of gravity at Earth’s surface, g, to the height that an object falls from, h, and the time it takes the object to reach the ground, t. Gravity acceleration at Earth’s surface has been measured many times. In British Imperial Units, Small Metric Units, and Large Metric Units, the standard values of g are: g = 32 feet per second-squared (ft/s2) g = 980 centimeters per second-squared (cm/s2) g = 9.8 meters per second-squared (m/s2). Theory Newton succeeded in explaining gravitational acceleration using his Laws of Motion and Universal Gravitational Force. According to Newton, F = ma, where F is the force on an object, m is the object’s mass and a is the acceleration of the object produced by the force. Newton also demonstrated that the gravitational force of attraction between two bodies, Fg, can be expressed as: M m Fg = G _____ , r2 where G is the Universal Gravitational Constant (6.67 X 10-11 N kg2/m2), M and m are the masses of the two objects and r is the center-to-center distance between the two objects. In the case where M = mass of the Earth, m = mass of an object on Earth’s surface and R is the Earth’s radius, we remember that the object’s weight is a force defined by the product mg, we can manipulate the two equations above to obtain: Mm M g = a = F/m = Fg /m = G _____ = G — . R2 R2 One way to measure g is to measure how much time (t) it takes for an object to fall to the surface after being dropped from a height h. This formula for this calculation, an early product of calculus, is: 2h g = ___ . t2 In the small metric system, height is in centimeters (cm) and time is in seconds. In the Large Metric System (also called the SI or MKS System), height is in meters (m) and time is in seconds. Procedure : If you were performing this experiment in an in-class laboratory, you would be given one or more small iron balls. These would be utilized with a variant of the Free-Fall apparatus shown in the graphic below. Free Fall Table Height h Time 1 Time 2 Time 3 Average time (tav) g= 2h/tav2 PE (cm) (sec) (sec) (sec) (sec) (cm/s2) ______ _____ _____ _____________ _________ ______ 40 0.27 0.26 0.27 50 0.33 0.31 0.34 60 0.36 0.35 0.37 80 0.40 0.41 0.41 100 0.43 0.45 0.44 _______________________________________________________________ gav = Question: The Earth’s radius is about 6400 km. The International Space Station (ISS) orbits about 400 km above Earth’s surface. So the center-to-center distance between Earth’s center and the space station is about 6800 km. Estimate Earth’s gravitational acceleration at the space station orbital height, giss. Use the ratio approach: giss / g = giss /980 = (6400 /6800)2 = giss = cm/s2.
According to legend, Galileo Galilei dropped two balls of different mass from the top of the leaning tower of Pisa in 1589. Whether or not this public experiment ever took place, Galileo was able to demonstrate that, contrary to Aristotle’s teaching, all bodies fall at the same rate regardless of mass, assuming that one is not so tenuous that it is slowed by air resistance.
In this experiment, an equation is presented relating the acceleration of gravity at Earth’s surface, g, to the height that an object falls from, h, and the time it takes the object to reach the ground, t.
Gravity acceleration at Earth’s surface has been measured many times. In British Imperial Units, Small Metric Units, and Large Metric Units, the standard values of g are:
g = 32 feet per second-squared (ft/s2)
g = 980 centimeters per second-squared (cm/s2)
g = 9.8 meters per second-squared (m/s2).
Theory
Newton succeeded in explaining gravitational acceleration using his Laws of Motion and Universal Gravitational Force. According to Newton,
F = ma,
where F is the force on an object, m is the object’s mass and a is the acceleration of the object produced by the force.
Newton also demonstrated that the gravitational force of attraction between two bodies, Fg, can be expressed as:
M m
Fg = G _____ ,
r2
where G is the Universal Gravitational Constant (6.67 X 10-11 N kg2/m2), M and m are the masses of the two objects and r is the center-to-center distance between the two objects.
In the case where M = mass of the Earth, m = mass of an object on Earth’s surface and R is the Earth’s radius, we remember that the object’s weight is a force defined by the product mg, we can manipulate the two equations above to obtain:
Mm M
g = a = F/m = Fg /m = G _____ = G — .
R2 R2
One way to measure g is to measure how much time (t) it takes for an object to fall to the surface after being dropped from a height h. This formula for this calculation, an early product of calculus, is:
2h
g = ___ .
t2
In the small metric system, height is in centimeters (cm) and time is in seconds. In the Large Metric System (also called the SI or MKS System), height is in meters (m) and time is in seconds.
Procedure :
If you were performing this experiment in an in-class laboratory, you would be given one or more small iron balls. These would be utilized with a variant of the Free-Fall apparatus shown in the graphic below.
Free Fall Table
Height h Time 1 Time 2 Time 3 Average time (tav) g= 2h/tav2 PE
(cm) (sec) (sec) (sec) (sec) (cm/s2)
______ _____ _____ _____________ _________ ______
40 0.27 0.26 0.27
50 0.33 0.31 0.34
60 0.36 0.35 0.37
80 0.40 0.41 0.41
100 0.43 0.45 0.44
_______________________________________________________________
gav =
Question: The Earth’s radius is about 6400 km. The International Space Station (ISS) orbits about 400 km above Earth’s surface. So the center-to-center distance between Earth’s center and the space station is about 6800 km. Estimate Earth’s gravitational acceleration at the space station orbital height, giss. Use the ratio approach:
giss / g = giss /980 = (6400 /6800)2 =
giss = cm/s2.
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