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Revision:AQA A2 Gravity

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[TSR Wiki]] > Study Help > Subjects and Revision > Revision Notes > Physics > A Level

Gravity

1 Newtons law of (universal) gravitation
2 Kepler's 3rd law of planetary motion
- 2.1 Calculating mass of Objects
- 2.2 comments

Newtons law of (universal) gravitation

Definition:

Two point masses will experience a force of attraction which is:

Proportional to the product of the masses
Inversely proportional to the square of their distances apart

$F \propto m_1m_2$

$F $\propto$ $\frac{1}{r^2}$$

What the defintion is saying is quite simple when you think about it. It says that the bigger the mass of an object the larger the force, and the further from the object the smaller the force becomes.

It is important to firstly learn this definition word for word and especially remember to include the points in bold. Gravity is a force of attraction, this must be mentioned whenever asked for the defintion and for the maths we assume all the masses are point masses, with distance, r, being measured from the centre of mass of the objects.

A point mass is the point where it can be assumed that all the mass is located, this means that distances are ALWAYS measured from the centre of mass of an object.

So when we combine these together we get

$F$\propto$ $\frac {m_1m_2}{r^2}$$

It is not an equation yet as it is missing a constant, G, which finishes the equation by taking into account the strength of gravity which is weak, especially when compared to the 3 other fundamental forces as well as satisfying the unit of force, the newton.

So finally we get:

$F = G \frac {m_1m_2}{r^2}$

G = UNIVERSAL GRAVITATIONAL CONSTANT = $6.67\times10^{-11} Nm^2kg^{-2}$

You can also write the formula as follows, with big M and little m. Big M is the mass of the bigger object ie the sun and little m is the smaller mass ie the Earth. In the A level you will always have a large mass with a smaller mass. Use M and m if you find this easier.

$F = G \frac {Mm}{r^2}$

Kepler's 3rd law of planetary motion

A common question in the A2 paper involves deriving from first principles Kepler's 3rd law. We do this my using our knowledge of both Newtons law of gravitation and uniform circular motion.

Kepler's 3rd law states:

$T^2 \propto r^3$

direction of force towards centre of sun

T = period of the planet

r = distance from the sun

We will work this out by using the example of the earth orbiting the sun.

First using circular motion we know that the centripetal force act TOWARDS the centre of a circle. We also know that centripetal force can always be desctibed as another force. In this case we know that the force which pulls the earth towards the sun is gravity.

This is are starting point.

$G\frac {m_1m_2}{r^2}=\frac {mv^2}{r}$

Next we divide through by m this is the mass of the ORBITING object in this case the earth, and multiply through by r to simply.

$G\frac {m_1}{r}={v^2}$

|NOTE: The velocity of the satellite orbiting is not dependent on its mass. This is very useful to know and later on it is this realisation which allows us to calculate the mass of objects just by looking at the distance a satellite is from the object,r, and the length of the period, T, of the satellite around the object.|

Now we need to use another bit of information we know about circular motion. We need to introduce the period, T into the equation. We do this by using the equation for linear velocity in circular motion.

We know that $v = \frac {2\pi r}{T}$

We substitute this in to give us:

$G \frac {m_1}{r} = ( \frac {2 \pi r}{T} )^2$

Multiply out the brackets and we get:

$G \frac {m_1}{r} = \frac {4 \pi^2 r^2}{T^2}$

Now we just need to get T^2 as the subject. Simply rearrange the formula:

$T^2 = \frac {4 \pi^2}{Gm_1}$r^3$$

And finally mention that $\frac {4 \pi^2}{Gm_1}$ is a constant, m1 is a constant as it refers to the mass of the sun.

$\therefore$ $T^2 \propto r^3$

Calculating mass of Objects

As you may have noticed from the above formula, as the velocity of an orbiting object is not dependent on its mass, we can calculate the mass of the object being orbited by knowing just the distance the objects are apart and the period, T, of the orbiting object. This is why we know the mass of the Earth to a very high level of accuracy, by looking at the distance of satellites and their period around the Earth.

As an example lets look at the mass of the sun.

First we rearrange the equation so M is the subject:

$T^2 = \frac {4 \pi^2}{Gm_1}$r^3$$

$M =\frac {4 \pi^2}{GT^2}$r^3$$