Gravitational Force vs. Centripetal Acceleration

Can someone please explain this conceptually: Suppose we have two probes, one one the surface of the earth and one in geosynchronous orbit. Since $a_c = \omega^2r$, the probe that is further out, i.e., the one in geosynchronous orbit, has a larger centripetal acceleration. But the probe that is further out also has a lower gravitational force pulling it inward, since $\mathbf F_g = \frac{GM_em}{r^2}$, or alternatively,the satellite further out has a lower acceleration due to gravity because that depends inversely on $r^2$. So could someone please explain how the probe further out can have both a higher centripetal acceleration yet a lower acceleration due to gravity than the probe on the surface of the Earth?

• Do gravitational force and acceleration force need to be equal or cancel or something? May 5, 2015 at 3:29
• You are forgetting the normal force exerted by the surface of the earth. May 5, 2015 at 3:50

The formula you've referenced, $\vec a_c = -\omega^2\vec r$, must define $\vec r$ as a function of the angular position, which depends on time (uniform motion and all). Something like $\vec r(\omega t) = r(\hat x \sin \omega t + \hat y \cos \omega t)$.

$\omega$ essentially represents the angular speed, which means for constant $\omega$ you have an orbital speed $v$ that depends on distance $r$ through the relation $a_c=\frac{v^2}r$. This applies to any uniform circular motion.

The force of gravity at a given distance gives you the acceleration, and with a suitable orbital speed the motion will be uniform. However, as you've noticed, the angular speed for uniform circular motion due to a force $\propto \frac1{r^2}$ (like gravity) will have decreasing angular speed with increasing $r$.

If you're only concerned with the magnitude of $a_c$, this can be shown as follows:

$$a_c = \omega^2r = \frac{v^2}r$$

First, you can observe the following:

$$v^2 = \omega^2r^2$$

$$v = \omega r$$

But you also need to be mindful that $a_c$ itself depends on $r^2$ via $a_c = \frac {\mathbf F_g} m = \frac{GM_e}{r^2}$.

So orbital speed depends on $r$, and angular speed depends on $r$, for uniform orbital motion due to gravity, but in different ways:

$$\sqrt \frac{GM_e}{r} = v$$

$$\sqrt \frac{GM_e}{r^3} = \omega$$

So in a way you're right, if the angular speed is kept constant, you need a larger $a_c$ with increasing $r$ to create uniform orbital (circular) motion. However this is only because the orbital speed increases with distance for constant angular speed $\omega$, which in turn requires a larger $a_c$ in order for the motion to still be uniform and circular. But under gravitation, $\omega$ is not constant because the force doesn't grow linearly with distance (it shrinks in fact), and $a_c$ is lower with increasing $r$ as a result.

Regarding your two probes, let's call them E (on earth) and S (in space)... Consider that E is not actually in uniform circular motion at all - it is on the surface of the earth, which is itself spinning. In fact, it is the other way around, probe E is moving too slow to achieve uniform circular motion around the earth, and would fall down.

Think of it this way - if the acceleration felt is larger than what is required for uniform circular motion at a given distance and speed (as is the case near the earth's surface for every-day situations), the object would approach the center, in other words fall to the ground. Consider how far and hard you have to throw something at ground level into the horizon for it to never hit the ground; geosynchronous / geostationary orbits are exactly this, but at a height where their required speed for uniform motion is small enough to match the speed at which the earth rotates.

It may help to try and work out the orbital speed (not in terms of $\omega$) for S, and then compare it to the orbital speed needed for uniform motion under gravity at smaller and smaller distances to the earth - you should find that as you come closer, you need to be moving considerably faster to maintain uniform circular motion ($v^2 \propto \frac1r$). And with any increase in speed, for the motion to remain uniform and circular at the same radius, the acceleration must increase ($v^2 \propto a_c$).

Another way to look at it is that in a geostationary orbit, the acceleration due to gravity and the centripetal acceleration are equal and exactly cancel each other out. This is true of ANY orbit, but with geostationary orbit of course the angular speed of the satellite exactly matches the rotation of the earth.

A probe in an orbit further out than geostationary will rotate more slowly because of the drop off in gravitational acceleration and a probe in an orbit closer to the earth will rotate more quickly because gravity is stronger. Otherwise either gravitational or centripetal acceleration will dominate and the probe will either fall to a lower orbit or climb to a higher orbit.

An object sitting stationary relative to the surface of the earth, however, is not in equilibrium. The acceleration due to gravity at the earth's surface is about 9.8 m/s2, whereas the centripetal acceleration is only 0.033 m/s2.