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Orbits >>
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3.3 - Newton's Law of Gravitation
Objectives:
- To understand the Law of Gravitation
- To be able to calculate the force between any two objects
- To understand the nature of the inverse square law
- To understand the concepts of weight and mass
- To be able to calculate the weight of an object
- To understand the concept of gravitational field strength
The basic physics: any two masses attract each other with a force that depends on the product of their mass and is inversely proportional to the distance that separates them. The constant of proportionality, know as the Universal Gravitational Constant, \(G\):
\[G = 6.67 \times 10^{-11}\,\text{Nm}^{2}\text{kg}^{-2}\]
\[G = 6.67 \times 10^{-11}\,\text{Nm}^{2}\text{kg}^{-2}\]
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Cavendish went to later to measure \(G\) and in effect then measured the mass of Earth and Sun.
Weight and Gravitational Field Strength
WEIGHT is the name given to the force that gravity pulls down on a specific object. The amount of 'stuff' in the object itself is called MASS. Mass can also be thought of its 'resistance' to a change in motion or its inertia. We usually denote weight by \(mg\). At the surface which is most of the time in physics problems(!), we tend to assume that has a fixed value of \(g = 9.81 \,\text{m/s}^2\) and think of it as the acceleration due to gravity. Actually it is the gravitational field strength, or how much force the Earth pulls on each kilogram of mass. Its units are \(\text{N/kg}\). In the absence of air resistance, a falling object accelerates at the same rate as the gravitational field strength. This is not a coincidence but rather a direct consequence of Newton's 2nd Law.
\[a=\frac{F}{m}\]
\[a=\frac{mg}{m}\]
\[a=g\]
\[a=\frac{F}{m}\]
\[a=\frac{mg}{m}\]
\[a=g\]
Another way to think of this is that mass does not change depending where you travel too, but the weight due to gravity does. An object in space may be weightless, but it still has mass. Kicking it will still hurt.
The gravitational field strength, g, varies as the force acting on an object varies. It is defined as the force due to gravity per kilogram of mass:
\[g = \frac{F}{m}\]
\[g = \frac{GM}{r^{2}}\]
The gravitational field strength, g, varies as the force acting on an object varies. It is defined as the force due to gravity per kilogram of mass:
\[g = \frac{F}{m}\]
\[g = \frac{GM}{r^{2}}\]
Substituting the known values for the radius of the Earth, the mass of the Earth and the Universal Gravitational Constant yields the expected value of \(g = 9.81 \,\text{N/kg}\) . The gravitational field strength varies from location to location and from planet to planet. Students often believe that gravity 'stops' in space, i.e. outside the atmosphere as they see astronauts floating in the ISS. Adding the space station's altitude of \(400 \,\text{km}\) to the radius results in a tiny reduction in the gravitational field strength.
Example 1 - Another Planet's Gravity MC Questions
A very common AP question is to compare the gravity of Earth to that of another that has (say) twice the mass and twice the radius of the Earth. A good technique is to use altered variables denoted by a prime, '. Note that \( G \) is universal and does not need to be changed.
\[g{}' = \frac{GM{}'}{r{}'^{2}}\]
\[g{}' = \frac{GM{}'}{r{}'^{2}}\]
Then substitute the changed values: \(M{}' = 2M\) (twice the mass of Earth) and \(r{}' = 2r\) (twice the radius of Earth). Note that I have used brackets to reduce potential errors!
\[g{}' =\frac{G(2M)}{\left ( 2r \right )^{2}}=\frac{2GM}{4r^{2}}\]
\[g{}' =\frac{G(2M)}{\left ( 2r \right )^{2}}=\frac{2GM}{4r^{2}}\]
As we know that \(\frac{GM}{r^2} = g_{Earth}\), this simplifies down to half that of Earth's:
\[g{}'=\frac{2GM}{4r^{2}}=\frac{2}{4}\left ( \frac{GM}{r^{2}} \right )=\frac{g}{2}\]
\[g{}' =\frac{g}{2}\]
\[g{}'=\frac{2GM}{4r^{2}}=\frac{2}{4}\left ( \frac{GM}{r^{2}} \right )=\frac{g}{2}\]
\[g{}' =\frac{g}{2}\]
Other Resources
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