4. Gravity and Orbits#
4.1. Playing with Newton’s Laws of Motion and Gravitation#
For any two objects, the force of gravity is directly proportional to the masses and indirectly proportional to the square of the distance between them. Mathematically, this is given by
where \(d\) represents the distance between them and \(M_1,\ M_2\) are the individual masses. The constant \(G\) represents the constant of universal gravitation, but in our cases it won’t be necessary to the actual value because we will be comparing force with some other standard.
4.1.1. Changing the Distance#
How would the gravitational force between the Earth and Moon change if the distance \(d\) between them was doubled?
We can compare the current Earth-Moon gravitational force \(F_g\) to the force \(F^\prime_g\) supposed in our hypothetical scenario using a ratio. From there, we can show that
In the above ratio, we see that the \(G M_\oplus M_{\rm Moon}\) factor is on top and bottom, which means that it simply cancels (or becomes unity). The factor \(d^2\) also cancels in the same way. However the factor \(2^2 = 4\) in the denominator remains. The result is
This means that the gravitational force in the hypothetical scenario \(F^\prime_g\) is only \(1/4\) of the original force \(F_g\). This should make sense because the Earth and Moon are \(2\times\) farther apart, where the strength of the force decreases by a factor of \(1/2^2\). Note: the force changed with the square when the distance changed.
4.1.2. Changing the Mass#
How would the gravitational force on you change if the Earth’s mass is doubled?
We can make a comparision using a ratio, where the new force is \(F^\prime_g\) and the original force is \(F_g\). From there, we have
In this case, the \(G M_\oplus M_{\rm you}\) and \(d\) factors cancel again. But there remains a factor of \(2\) in the numerator. The result is
This means that the gravitational force on you simply increases doubled with the increase in Earth’s mass. Note: the force changed proportionally when the mass was changed.
4.1.3. Gravitational Acceleration#
We can think about the gravitational force the Earth exerts on an object through: Newton’s 2nd law or the universal law of gravitation.
Through Newton’s 2nd law, we have \(F = m\times a = ma\). When applying this formula to the gravitational force, we often write \(F_g = mg\). The \(g\) represents the acceleration due to gravity at Earth’s surface.
Earth has a mass \(M_\oplus\) and radius \(R_\oplus\), where the gravitational force on another object with mass \(m\) using the universal law of gravitation is written as
In this case the multiplication is commutative, which means we can rearrange the order or regroup the terms. In this way, we find
Let’s replace the \(\frac{G M_\oplus}{R_\oplus^2}\) factor with \(g\), or
We recovered the formula from Newton’s 2nd law. This is a consequence of the equivalence principle if you would like to know more.
What does this mean?
All objects (near the Earth) fall at the same rate! This is independent of the mass. See the video below that demonstrates that a bowling ball and feather fall at the same rate, when in vacuum.
4.2. Circular and Escape Velocity#
4.2.1. Circular Velocity#
The velocity \(v\) of an object is basically defined by ratio of the distance \(d\) it has traveled with the travel time \(t\), or
More precisely, the velocity includes direction (e.g., to the left, right, up, or down), which is encoded using at sign (\(+\) or \(-\)). But this is beyond the purposes of the basic definition here. From Motion of Astronomical Bodies, we showed that acceleration is the change in velocity over time, or
Let’s modify the form of the top equation so that it is \(t = d/v\), using algebra. Then substitute it into the equation for acceleration. (See Appendix 8 from 21st Century Astronomy for a full description of the reasoning here.) Through this substitution, we find that
The above expression represents the centripetal acceleration \(a_c\) (or acceleration from circular motion). The circular velocity \(v_c\) is defined as the speed of an object in a circular orbit with a radius \(r\), where we determine this speed by setting the centripetal force \(F_c = ma_c\) equal to the gravitational force \(F_g\). Mathematically, this is given as
Notice that we can cancel a factor of \(\frac{m}{r}\) from each side, which produces
where \(M\) represents the mass of the central object (e.g., Sun, Earth, \(\ldots\)) that the object with mass \(m\) is orbiting. The distance traveled by the object is simply the radius \(r\) of the orbit. Finally, we have the constant of universal gravitation \(G\).
An Earth radius \(R_\oplus\) is defined as \(6371\ {\rm km}\), which is a little smaller than the value determined via the circumference because the Earth’s radius varies with latitude due to its spin. This value is the accepted mean radius between the value at the equator and that at the pole. But, let’s define the circular velocity at Earth’s surface \(v_c^{\rm surf}\), which would be
This is the speed required to keep an object moving along Earth’s surface in a circular path (neglecting air resistance, buildings, trees, \(\ldots\)). As the object moved horizontally, Earth’s surface would be falling away because it’s curved. However, gravity would exert a force on the object that changes its path from a straight line and the trajectory becomes a circle instead.
Cannonballs and other objects don’t move \(7.9\ {\rm km/s}\) (or ~28 kph), but rockets can routinely reach that speed to put satellites (and people) into orbit. Let’s see how the circular velocity (at the surface) on the Moon compares to the value on Earth. For this, we can determine the ratio of the two velocities as
Written this way, we can easily calculate the ratio if we know the Moon’s mass and radius relative to the Earth. For this, we can consult the NASA Moon Fact Sheet to get \(M_{\rm Moon}/M_\oplus = 0.0123\) and \(R_{\rm Moon}/R_\oplus = 0.2727\). Then we can find
Therefore the circular velocity at the Moon’s surface \(v_c^{\rm Moon}\) is \({\sim}21.2\%\) of the circular velocity at Earth’s surface \(v_c^{\rm Earth}\), or \(1.67\ {\rm km/s}\).
Does this make sense?
Yes! The Moon is both less massive and smaller than the Earth. The mass ratio is \({\sim}1/80\) as compared the the radius ratio \({\sim}1/4\), so the mass decreases a lot faster than the radius. This tells us that we should expect a lower circular velocity.
Moreover, it should give us insight into why we want to have Moon bases/gateways for Solar System exploration. Getting to a lunar orbit (orbiting around the Moon) requires less velocity than getting to an equivalent Earth orbit (e.g., LEO). Less velocity translates into less fuel required!
4.2.2. Escape Velocity#
Sending a spacecraft to another planet (e.g., Mars) requires launching with a speed that is greater than the escape velocity \(v_{\rm esc}\). The escape velocity is determined as the speed which all of the gravitational potential energy is transformed into kinetic energy. We can determine this by setting the two energy equal to each other, or
and solving for \(v_{\rm esc}\) (by canceling the \(m\) from each side) as
Notice that this equation looks very similar to the formula we found for the circular velocity. In fact, we can show how they are related by using the ratio
where all the factors cancel except for a \(\sqrt{2}\). This tells us that \(v_{\rm esc} = \sqrt{2}v_c\), or the escape velocity is \({\sim}1.4\times\) greater than the circular velocity at a given radius \(r\).
To leave the Earth, a rocket must have a speed of \(\sqrt{2} \times 7.9\ {\rm km/s} = 11\ {\rm km/s}\).
If we know the relative circular velocity on another body (i.e., planet, moon, asteroid), then we can use the same ratio to determine the relative escape velocity because the factor of \(\sqrt{2}\) occurs in both the numerator and denominator (i.e., cancels). Below is the ratio of circular velocities, with the Moon subscripts removed.
4.2.3. Center of Mass#
When two objects are closer to having the same mass (e.g., the stellar binary Alpha Centuri AB), both objects experience an acceleration due to one another’s gravity. This changes their path from a straight line so that the fall along a curve with a common point called the center of mass.
4.3. Calculating Mass from Orbital Periods#
4.3.1. Newton’s Version of Kepler’s Third Law#
The orbital (sidereal) period \(P\) is the time for a planet to return to the same location in space, or complete one orbit. Let’s assume a circular orbit for simplicity. The distance traveled along the orbit is the circumference of a circle, or \(d = 2\pi r\). In this case the circle’s radius is the semimajor axis \(a\), where \(d = 2\pi a\).
Recall the formula for the circular velocity, where we also substitute \(r = a\) to get
Then we can determine the orbital period \(P\) using the basic definition for velocity, or
It would be easier if we squared both sides to remove the square root, which gives
This is Newton’s version of Kepler’s 3rd law, which includes the central mass \(M\). When Kepler empircally determined his 3rd law, he was using the Solar System planets, which coincidently in that case \(4\pi^2/(GM) \approx 1\) and Kepler’s form is recovered (\(P^2 = a^3\)). Recall that Kepler used time units of years and distance units of AU.
Using the same units as Kepler, but also introducing the mass unit: solar mass, we can write this version of Kepler’s 3rd law in a simpler form:
Remember the units are years, AU, and solar masses for this formula. You must convert to years, AU, and solar masses, if the given values have different units (e.g., days, meters, kg).
4.3.2. Mass of the Jupiter#
If we can measure the size and period of any orbit, then Newton’s version of Kepler’s 3rd law can be used to measure the mass of the central body. To do so, we just need to re-arrange the formula so that the mass \(M\) is the dependent variable:
Suppose that we observe the Jupiter’s moon Europa over a month. We find that it takes \(3.55\ {\rm days}\) for Europa to completely orbit Jupiter and return to the same position. Also, we measure that Europa is \(4.48 \times 10^{-3}\ {\rm AU}\) from Jupiter on average. From these measurements, it would be easier if we converted the orbital period to years.
Then we can apply our simpler formula directly to measure the mass of Jupiter. We find,
We found that Jupiter’s mass is \(9.52 \times 10^{-4}\ M_\odot\), where the true value is \(9.54 \times 10^{-4}\ M_\odot\). Below is a python code that performs the same calculation.
P_meas = 3.55 #measured orbital period of Europa (in days)
P_yr = 3.55/365.25 #converted period to years
a_meas = 4.48e-3 #measured semimajor axis (in AU)
M = a_meas**3/P_yr**2
print("The mass of Jupiter (in solar masses) is %1.2e M_sun." % M) #The notation "e-04" is the same as "times 10^{-4}"
The mass of Jupiter (in solar masses) is 9.52e-04 M_sun.
4.4. Tidal Forces#
The strength of the gravitational force depends on the mass and the square of the distance between them. However, the strength of the tidal force depends on the mass and the cube of the distance between them. Therefore the tidal force acts over much shorter ranges because as the distance increases the tidal force strength decreases rapidly.
The equation of the tidal force comes from the difference of the gravitational force on one side of a body compared with the other side. We can approximate the tidal force of the Moon acting on a mass on Earth’s surface by
If we combined the denominator, we can see that it does indeed involve the cube of the distance. The middle term looks like a familiar form of gravity that we’ve see before. The leading 2 comes from how we are comparing the force from each side of the Moon that is acting on \(m\).
We can write a similar equation for the Sun’s tidal force acting on the same mass \(m\) on the Earth. The Sun’s tidal force is not negligible because it is so much more massive, although it is a lot farther away. This tidal force from the Sun is
Given these two tidal forces, we can compare their strengths throught the ratio
which looks complicated, but many of the terms cancel. The remaining terms form the relation:
This still seems a little complicated when we want to substitute values, but we only need to convert the Moon’s mass (in Earth masses) that we saw before into solar masses. To do this, we use a conversion factor
using the definition of an Earth mass. The Moon’s semimajor axis is already measured to be \(0.00257\ \rm AU\). Now we can begin plugging in some values:
The tidal force betwen the Earth and Moon is \(2.18\times\) stronger than the tidal force between the Earth and Sun. This is why the Moon’s tide affects the Earth more than the Sun’s tide. However the Sun’s tide is not neglibible!
Below is a python script that demonstates the calculation.
M_moon = 0.0123 #Moon's mass (in Earth masses)
M_moon_sol = M_moon/3.33e5 #convert the Moon's mass into solar masses
a_EM = 0.00257 # semimajor axis of the Moon's orbit
tide_ratio = M_moon_sol/a_EM**3
print("The Moon's tide on the Earth is %1.2f times stronger than the Sun's tide." % tide_ratio)
The Moon's tide on the Earth is 2.18 times stronger than the Sun's tide.