8.2.1 State Newton’s law of universal gravitation

Newton hypothesized that any two objects with mass will attract each other. The force of attraction is gravity, the same force that holds us on the ground and makes things falls. Newton supposed that gravity acts everywhere and can be described by the following equation:

\begin{align} F = G {M_1 M_2 \over r^2} \end{align}

Where M1 and M2 are the masses of the two objects, r is the distance between the masses (masses must be point masses) and G is the universal gravitational constant. Note: for spheres r is the distance between the center of their masses.

\begin{align} G = 6.67 \times 10^-11 Nm^2kg^-2 \end{align}

Newton’s law of universal gravitation is very similar in form to Coulomb’s Law (which deals with electric charges), however gravity is always attractive. There are positive and negative electric charges, but no one has ever observed “negative mass.”

8.2.2 Define gravitational field strength
8.2.3 Derive an expression for the gravitational field as a function of distance from a point mass.

In physics we often talk about fields (electric, magnetic, gravitational…). Fields are an abstract concept. Tsokos describes a field as, “…[A] field in this context, means that at every point in space a certain quantity is defined.” For gravity this means that if we put a mass in a gravitational field the mass will feel a force exerted on it.

Gravitational field strength can be defined as the ratio of the force a mass would experience to the mass itself. Thus the gravitational field strength can be defined as:

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

Recall that weight is the force of gravity acting on a mass. The weight of an object can be found using Newton’s 2nd law:

\begin{equation} F = mg \end{equation}

Where g is the acceleration due to gravity (at the location of the measurement). Therefore:

\begin{align} mg = G {Mm \over r^2} \end{align}
\begin{align} g = G{M \over r^2} \end{align}

We can see that the gravitational field strength is simply the acceleration of gravity at a given location. The gravitation field strength is a vector quantity (as are most fields), it points in the direction that a mass would accelerate if placed in the field.

Note: The subscripts 1 and 2 or lower case “m” are simply labels and not in themselves important.

8.2.4 Derive an expression for gravitational field at the surface of a planet.

In the case of the Earth (or other planets), M is the mass of the earth (or other planet) and r is the distance from the center of the planet. If the area of interest is the surface of the planet, r is the radius of the planet.

\begin{align} g_p = G {M_p \over r_p^2} \end{align}

The "gravitional field" is a way of visualizing either the acceleration or force due to a massive body. Gravity is always attractive and thus the "gravitional field" for a single massive body looks like the picture below:


Stolen from Tsokos 2nd Edition page 288

8.2.5 Solve problems involving gravitational forces and fields.

Example 1: Find the average force between the sun and the earth.

Example 2: If the distance between two bodies is doubled what happens to the gravitational force between them?

Example 3: Find the acceleration due to gravity on a planet 10 times as massive as the earth and with a radius 20 times as large.

8.2.6 Define gravitational potential energy and gravitational potential

In the past we have defined gravitational potential energy as , where m is the mass of the object, g is the acceleration of gravity and h is the height above the defined (choosen) zero potential energy point. This is valid only when the acceleration due to gravity, or gravitational field strength, is constant.

Take the example of a mass m a distance R from mass M, as shown below.


Tsokos 2nd Ed. Pg 289

If the mass is moved the work done on the mass is equal to the change in potential energy (assuming the movement is slow so as not to add kinetic energy). Work is defined as , however the force is not constant and varies with distance. To find an express for work we need calculus, the derivation is not IB required information, I provide it simply for those who are interested.

\begin{align} W = F \bullet d \end{align}
\begin{align} W = G {Mm \over r^2} \times \Delta d \end{align}
\begin{align} W = - {\int_R^\infty} G {Mm \over r^2} dr = G {Mm \over r} \big |_R^\inf{} \end{align}

Note: That the work done is negative. This is because the force and displacement are in opposite directions. This also makes gravitational potential energy negative.

From this we can say that two objects with mass separated by a distance R have potential energy:

\begin{align} PE_g = -G {m_1 m_2 \over R} \end{align}

If the mass is moved very far away, to infinity, the mass will have zero gravitational potential energy!

The potential energy is negative, this is because we have defined the potential energy to be zero when the objects are infinitely far apart (which makes sense). The work done, or change in potential energy is completely independent of the path taken, i.e. it only depends on the initial and final position. This is true when dealing with conservative forces (gravity, electricity, magnetism…).

In the diagram to the right, a mass m it moved along the path shown (the big green dot is a planet with mass M) the work done is:

\begin{align} W = -G {Mm \over R} \end{align}
8.2.7 State the expression for gravitational potential due to a point mass.

In physics we talk about potentials, this should not be confused with potential energy, the two are related but are two different things.

The gravitational potential is a field but it is a scalar not a vector. One way of thinking of the gravitational potential is that it is the gravitational potential energy per unit mass. The gravitational potential due to a point mass (or uniform sphere) can be described mathematically as:

\begin{align} V = - {E_p \over m} = - G{M \over r} \end{align}

Where V is the gravitational potential. The gravitational potential gives us a way to describe “gravity” due to one mass. We are able to attain some information about gravity with out having to use or know about a second mass.

In higher level physics the concept of the potential can be very useful, the gradient (mathematical operation) of the gravitational potential is equal to the negative of the gravitational field strength.

8.2.8 Explain the concept of escape velocity.

When a mass is thrown up, with a low initial velocity, it falls back down. From an energy argument, this could be said to be a result of the object having a non-zero (or negative) potential energy, there is potential energy that tries to make the mass fall back down. But if an object could have zero gravitational potential energy it would never fall back down. The total energy of a mass m, where the only force acting is gravity due to a second mass M, can be described as:

\begin{align} E = \frac {1}{2} mv^2 -G{Mm \over r} \end{align}

In this equation we assume that mass M is stationary.

If we want to send a satellite out beyond the earth or beyond the solar system, for exploration or research, we basically need to be able to throw it up with out it coming back down. In order for something to not come back down it needs to be “free of gravity.” As mentioned earlier if an object is moved to "infinity" it will have no gravitational potential energy, i.e. it is free of the effects of gravity. The force of gravity has no limits to how far it reaches, being free of gravity means that an object’s kinetic (assuming no other forces) energy is larger than it gravitational potential energy.

Since gravitational potential energy is always non-positive (zero or negative), this means that for an object to be free of gravity it must have either zero or positive total energy. In an equation this can be written as:

\begin{align} E = \frac{1}{2} mv_o^2 - G{Mm \over r} = \frac{1}{2}mv_{\infty}^2 \geq 0 \end{align}

Where $v_0$is the objects initial velocity and $v_\infty$ is the objects velocity once it reaches "infinity"

8.2.9 Derive an express for the escape velocity from a planet.

If an object is to just barely escape gravity, that is to get to "infinity" and have no velocity, the total energy must be equal to zero. If we then solve for the initial velocity:

\begin{align} \frac{1}{2} mv_{escape}^2 - G{Mm \over r} = 0 \end{align}
\begin{align} v_{escape} = \sqrt{G{2M \over r}} \end{align}

This velocity is known as the escape velocity, it is the minimum velocity for an object to escape the gravitational pull of another mass M. Note that the escape velocity is independent of the objects mass. The escape velocity for the earth is approximately 11 kms-1.

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