Gravitational Force: Difference between revisions
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In this problem, we are given the value of two masses m<sub>1</sub> and m<sub>2</sub> as 5.98 x 10<sup>24</sup> kg and 70 kg respectively and the distance is also given as 6.39 x 10<sup>6</sup> m. We substitute these values into the universal gravitational equation and solve for F<sub>grav</sub>.The solution is as follows: | In this problem, we are given the value of two masses m<sub>1</sub> and m<sub>2</sub> as 5.98 x 10<sup>24</sup> kg and 70 kg respectively and the distance is also given as 6.39 x 10<sup>6</sup> m. We substitute these values into the universal gravitational equation and solve for F<sub>grav</sub>.The solution is as follows: | ||
''F<sub>grav</sub>''<math> = G \frac{m_1 m_2}{r^2}\ </math> | |||
===Middling=== | ===Middling=== | ||
Revision as of 18:43, 27 November 2015
Main Idea
A Mathematical Model
What are the mathematical equations that allow us to model this topic. For example [math]\displaystyle{ {\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net} }[/math] where p is the momentum of the system and F is the net force from the surroundings.
A Computational Model
How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript
Units
In Si Unit, Gravitational Force F is measured in Newtons (N), the two masses, m1 and m2 are measures in kilograms (Kg), the distance is measured in meters (m), and the gravitational constant G is measured in N m2/ kg−2 and has a value of 6.674×10−11 N m2/ kg−2. The Gravitational Constant G have different values for different units. The value of constant G appeared in Newton's law of universal gravitation, but it was not measured until seventy two years after Newton's death by Henry Cavendish with his Cavendish experiment in 1798. The value of gravitational constant was also the first test of Newton's law between two masses in Laboratory.
Examples
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Simple
Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is in an airplane at 40,000 feet above earth's surface. This would place the student a distance of 6.39 x 106 m from earth's center.
In this problem, we are given the value of two masses m1 and m2 as 5.98 x 1024 kg and 70 kg respectively and the distance is also given as 6.39 x 106 m. We substitute these values into the universal gravitational equation and solve for Fgrav.The solution is as follows:
Fgrav[math]\displaystyle{ = G \frac{m_1 m_2}{r^2}\ }[/math]
Middling
Difficult
Connectedness
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History
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See also
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Further reading
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