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2011-05-27T05:31:05Z

Created page with 'A satellite on its way to the Moon is attracted towards the Earth with a gravitational force of <math>F_1</math>. The force gradually decreases as the distance, <math>x</math>, b…'

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A satellite on its way to the Moon is attracted towards the Earth with a gravitational force of <math>F_1</math>. The force gradually decreases as the distance, <math>x</math>, between the Earth and satellite increases. At the same time the Moon attracts the satellite with a gravitational force <math>F_2</math> (which increases as the distance to the Moon decreases.) These two forces oppose one other (see the diagram below). 
[[image:N_APP_EX_40_SOLN.png|TOP]] 

The distance between the Earth and the Moon is <math>3.810 \times 10^8</math> m. When the satellite is at a distance <math>x</math> from the Earth then it is at a distance <math>([3.810 \times 10^8] - x)</math> from the Moon. We are looking for a distance <math>x</math> at which <math>F_1 + F_2 = 0</math>, which happens when the ''magnitudes'' of these forces are equal. The gravitational force <math>F_1</math> depends on the mass <math>m_E</math> of the Earth, the mass <math>m_s</math> of the satellite, and the distance <math>x</math> between them; the force <math>F_2</math> depends on the mass <math>m_M</math> of the Moon , the mass <math>m_s</math> of the satellite, and the distance <math>([3.810 \times 10^8] - x)</math>. We will now write the equation <math>F_1 = F_2</math> and replace both <math>F_1</math> and <math>F_2</math> by appropriate formulas in the Newton Law of gravitation.

<math>{G m_E m_s \over x^2} = {G m_M m_s \over ([3.810 \times 10^8] - x)^2}</math>
 
Given that the Earth is 81 times more massive than the Moon, we can write: <math>m_E = 81 m_M</math>. Upon substitution into the above, we get:

<math>{G (81 m_M) m_s \over x^2} = {G m_M m_s \over ([3.810 \times 10^8] - x)^2}</math>
 
The constants <math>G</math>, <math>m_M</math> and <math>m_s</math> cancel out to leave us with:

<math>{81 \over x^2} = {1 \over ([3.810 \times 10^8] - x)^2}</math>
 
By taking the positive square root on either side of this equation, we get

<math>{9 \over x} = {1 \over (3.810 \times 10^8) - x}</math>
 
and after cross multiplication and isolation of <math>x</math>, we see that <math>x = 3.4 \times 10^8</math> m.

Newtons Laws EX 40 - Revision history

imported>Patrick: Created page with 'A satellite on its way to the Moon is attracted towards the Earth with a gravitational force of F_1. The force gradually decreases as the distance, x, b…'

imported>Patrick: Created page with 'A satellite on its way to the Moon is attracted towards the Earth with a gravitational force of $F_1$ . The force gradually decreases as the distance, $x$ , b…'