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Newton’s Law of Universal Gravitation

Newton’s Law of Universal Gravitation

Newton's Law of Universal Gravitation

Definition: Newton’s Law of Universal Gravitation

Every point mass attracts every other point mass by a force directed along the line connecting the two. This force is proportional to the product of the masses and inversely proportional to the square of the distance between them.

The magnitude of the attractive gravitational force between the two point masses, F is given by: \[F=G\frac{m_1m_2}{d^2}\]

where:\(F\) is in newtons (N), \(G\) is the gravitational constant \(\text{6.67} \times \text{10}^{-\text{11}}\) \(\text{N·m$^{2}$·kg$^{-2}$}\), \({m}_{1}\) is the mass of the first point mass in kilograms (kg), \({m}_{2}\) is the mass of the second point mass in kilograms (kg) and \(d\) is the distance between the two point masses in metres (m). For any large objects (not point masses) we use the distance from the centre of the object(s) to do the calculation. This is very important when dealing with very large objects like planets. The distance from the centre of the planet and from the surface of the planet differ by a large amount. Remember that this is a force of attraction and should be described by a vector. We use Newton’s law of universal gravitation to determine the magnitude of the force and then analyse the problem to determine thedirection.

For example, consider a man of mass \(\text{80}\) \(\text{kg}\) standing \(\text{10}\) \(\text{m}\) from a woman with a mass of \(\text{65}\) \(\text{kg}\). The attractive gravitational force between them would be: \begin{align*} F&=G\frac{m_1m_2}{d^2} \\ &= \left(\text{6.67} \times \text{10}^{-\text{11}}\right)\left( \frac{(\text{80})(\text{65})}{(\text{10})^2} \right) \\ &= \text{3.47} \times \text{10}^{-\text{9}}\text{ N} \end{align*}

If the man and woman move to \(\text{1}\) \(\text{m}\) apart, then the force is: \begin{align*} F&=G\frac{m_1m_2}{d^2} \\ &= \left(\text{6.67} \times \text{10}^{-\text{11}}\right)\left( \frac{(\text{80})(\text{65})}{(\text{1})^2} \right) \\ &= \text{3.47} \times \text{10}^{-\text{7}}\text{ N} \end{align*}

As you can see, these forces are very small.

Now consider the gravitational force between the Earth and the Moon. The mass of the Earth is \(\text{5.98} \times \text{10}^{\text{24}}\) \(\text{kg}\), the mass of the Moon is \(\text{7.35} \times \text{10}^{\text{22}}\) \(\text{kg}\) and the Earth and Moon are \(\text{3.8} \times \text{10}^{\text{8}}\) \(\text{m}\) apart. The gravitational force between the Earth and Moon is: \begin{align*} F&=G\frac{m_1m_2}{d^2} \\ &= \left(\text{6,67} \times \text{10}^{-\text{11}}\right)\left( \frac{(\text{5,98} \times \text{10}^{\text{24}})(\text{7,35} \times \text{10}^{\text{22}})}{(\text{0,38} \times \text{10}^{\text{9}})^2} \right) \\ &= \text{2,03} \times \text{10}^{\text{20}}\text{ N} \end{align*}

From this example you can see that the force is very large.

These two examples demonstrate that the greater the masses, the greater the force between them. The \(1/{d}^{2}\) factor tells us that the distance between the two bodies plays a role as well. The closer two bodies are, thestronger the gravitational force between them is. We feel the gravitational attraction of the Earth most at the surface since that is the closest we can get to it, but if we were in outer-space, we would barely feel the effect of the Earth’s gravity!

Remember that \(\vec{F}=m\vec{a}\) which means that every object on Earth feels the same gravitational acceleration! That means whether you drop a pen or a book (from the same height), they will both take the same length of time to hit the ground… in fact they will be head to head for the entire fall if you drop them at thesame time. We can show this easily by using Newton’s second law and the equation for the gravitational force. The force between the Earth (which has the mass \({M}_{\text{Earth}}\)) and an object of mass \({m}_{o}\) is

\(F=\dfrac{G{m}_{o}{M}_{\text{Earth}}}{{d}^{2}}\)

and the acceleration of an object of mass \({m}_{o}\) (in terms of the force acting on it) is

\({a}_{o}=\dfrac{F}{{m}_{o}}\)

So we equate them and we find that \[a_o = G\frac{M_{Earth}}{d_{Earth}^2}\]

Since it doesn’t depend on the mass of the object, \({m}_{o}\), the acceleration on a body (due to the Earth’s gravity) does not depend on the mass of the body. Thus all objects experience the same gravitational acceleration. The force on different bodies will be different but the acceleration will be the same. Due to the fact that this acceleration caused by gravity is the same on all objects we label it differently, instead of using \(a\) we use \(g_{Earth}\) which we call the gravitational acceleration and it has a magnitude of approximately \(\text{9.8}\) \(\text{m·s$^{-2}$}\).

The fact that gravitational acceleration is independent of the mass of the object holds for any planet, not just Earth, but each planet will have a different magnitude of gravitational acceleration.

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