Orbital Motion

8.3.1 State that gravitation provides the centripetal force for circular orbital motion.

If you attach a ball to the end of a string and swing the ball around the ball moves in a circle, no surprise. The string provides a centripetal force to keep the ball moving in a circle as opposed to moving in a straight line (Newton’s 1st Law). For a planet orbiting a star or a satellite orbiting the earth, gravity provides this centripetal force.

Assuming that the path of the satellite is circular we can use the Newton’s law of Universal Gravitation:

\begin{align} F = G {Mm \over r^2} \end{align}

and Newton’s 2nd law:

\begin{align} F = ma = m{v^2 \over r} \end{align}

to calculate the speed of an orbiting satellite:

\begin{align} G {Mm \over r^2}=m{v^2 \over r} \end{align}
\begin{align} v^2 = G{M \over r} \end{align}
\begin{align} v = \sqrt{G{M \over r}} \end{align}

Note that the velocity is independent of the mass of the satellite. The above equation will work for any planet or star (or any an object that has something in orbit) with mass M and radius r. If we want to calculate the velocity for a object orbiting the earth we can use the knowledge:

\begin{align} g=G{M_e \over R^2_e} \end{align}

Therefore we can calculate the velocity as:

\begin{align} v = \sqrt{gR_e} \end{align}

For an object orbiting just above the surface of the earth (ignoring air resistance and mountains) this comes out to be a speed of about 7.9 km/s.

8.3.2 State Kepler’s 3rd Law

After collecting data on the orbits of planets, Kepler theorized that that the period T of an orbiting planet is related to the radius R of the orbit by the following mathematical law:

\begin{align} T^2 \propto R^3 \end{align}

8.3.3 Derive Kepler’s 3rd Law

Kepler made this assumption with no knowledge of the Law of Universal Gravitation. Kepler’s 3rd Law can be easily derived, if T is the period (time for one revolution) then we can say:

\begin{align} v={2 \pi R \over T} \end{align}

Where R is the radius of the orbit. Using the result we found for orbital velocity above:

\begin{align} \left (2 \pi R \over T \right )^2 = {GM \over R} \end{align}
\begin{align} 4 \pi^2 R^3 = GMT^2 \end{align}
\begin{align} T^2 = {4 \pi^2 \over GM}R^3 \end{align}
\begin{align} T^2 \propto R^3 \end{align}

In the IB Data booklet Kepler’s 3rd law is written as:

\begin{align} {T^2 \over R^3} = constant \end{align}

8.3.4 Derive expressions for the kinetic, potential and total energy of an orbiting satellite.

8.3.5 Draw graphs showing the variation of the kinetic energy, gravitational potential energy and total energy with orbital radius.

The total energy for a orbiting satellite is equal to the kinetic energy plus the potential energy:

\begin{equation} E_{Tot} = E_k +E_p \end{equation}

We know from previous discussions that the gravitational potential energy is described by the equation:

\begin{align} E_p = - G{Mm \over R} \end{align}

The kinetic energy is equal to:

\begin{align} E_k = \frac{1}{2} mv^2 \end{align}

But we know that the speed of a satellite is dependent on its distant from the center of the object it is orbiting. We can therefore rewrite the kinetic energy as:

\begin{align} v^2 = G{M \over R} \end{align}
\begin{align} E_k = \frac{1}{2} m \left(G{M \over R} \right) \end{align}

Therefore we write the total energy as:

\begin{align} E_{Tot} = \frac{1}{2}G{Mm \over R} - G{Mm \over R}=-\frac{1}{2}G{Mm \over R} \end{align}

We can graph the results:


In the case of the object orbiting the earth the total energy is written as:

\begin{align} E_{Tot} = - \frac{1}{2}G{M_E m \over R_E} \end{align}

Notice that the total energy of the orbiting satellite is negative. Why? If the total energy was zero or positive the satellite would have enough energy to escape the pull of gravity or go to infinity. This would mean that the object has too much energy and would not orbit, i.e. the pull of gravity is not strong enough to keep the object moving in a circle.

8.3.6 Discuss the concept of weightlessness in both orbital motion and free fall

What does it mean to be weightless? Maybe a better first question is what does it mean to be weightful?

If you stand on a scale why does the scale register a reading? There are two sets of action and reaction force, one between you and the scale and another between the scale and the ground. When you stand on the scale the ground pushes up on the scale that allows the scale to push up on you and complete the action reaction pair. If this action reaction pair doesn’t exist the scale can’t give a reading.

During free fall the “ground” or surface that a scale would rest on is accelerating at the same rate as the scale, therefore the surface can not exert a force on the scale and therefore the scale would read zero. This is what is meant by weightlessness. You are not free of gravity, you simply can not measure your weight.

During orbital motion an object is continuously in free fall, it’s falling around the planet or whatever object it is orbiting. So when orbiting an object is weightless.

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