# Does trajectory of an object orbiting a planet depend on the object's mass? (With hypothetical Apollo example)

For example (hypothetically), if an Apollo crew of one of the lunar missions went on EVA just after Trans Earth Injection burn and carefully released a small bolt, or a plastic package from their food, or the famous feather near the CSM and then hold the spacecraft in the same attitude and didn't perform any course correction maneuvers, would this released object follow exactly same trajectory (as well as velocity and acceleration profile in time) as the heavier spacecraft, i.e. would it stay at same distance relative to CSM up until 0.05g event?

Sorry for asking such a simple question, I'm new to orbital mechanics and just trying to have a feel of its peculiarities on simple examples.

• If both objects are travelling at a speed sufficient to overcome the gravity of the body it is orbiting, then they are both in "free-fall" technically and will not move away from each other unless a force is applied to alter the velocity vector of either. Mascons, Drag, Debris, Solar Pressure are all additional factors that would, over time, move your objects apart. Think of an orbit as a free-fall but you're just moving in a perpendicular direction enough to "repeatedly miss the ground" at 9.8m/s^2. All the rules of the feather-bowling ball experiment still would apply. Nov 20 '19 at 14:11
• Perhaps worth noting that you don't even need to do an EVA to test this. Any object floating inside a spacecraft is doing so because it has the same orbital trajectory as the spacecraft itself, so there's no relative movement between the two. Nov 20 '19 at 15:38
• @NuclearWang there is a GREAT video of the station doing a reboost. While it does the reboost an astronaut is filiming a wrench inside the ISS. As the ISS boosts the wrench moves against the thrust! Otherwise its stationary. Ill see if I can find it. Nov 20 '19 at 15:59
• @NuclearWang The orbital trajectory isn't precisely the same; over time, you would see all of the "floating" objects get stuck on one of the walls. Of course, if there's air inside, the motion and resistance of air is going to be a much bigger effect. Even very small differences in orbital distance/velocity add up relatively quickly - there is a tension on e.g. the ISS due to this, and if you separated the ISS modules, they would very slowly drift apart. Nov 21 '19 at 8:38

Approximately, yes. The gross gravitational effects on the trajectories of the spacecraft and the other object will be the same.

The force of gravity between two objects is proportional to the product of their masses; by $$F = m a$$, the acceleration of each object cancels out its own mass ( $$a = \frac {F} {m}$$ ) and so depends on the mass of the other object.

Since the bolt is in a very slightly different location relative to the earth and moon during the trip, however, the influence of gravity on it will be very slightly different in amount and direction, so it won’t follow an exactly parallel trajectory. It will be extremely close, however, and in practice I believe the difference in gravitational influence will be tiny compared with the difficulty of releasing the object with exactly zero initial velocity relative to the spacecraft. There would also be other confounding factors: solar pressure on the two bodies would push them off course by different amounts, the spacecraft will be venting various things that will push it around, etc.

• If the object was placed directly "behind" the object then (subject to confounding factors) it should follow the same trajectory, which would mean it would stay the same time behind the vehicle, but that distance would vary as the velocity changed
– user20636
Nov 20 '19 at 16:14
• @JCRM Except for the fact that the relative Earth-moon positions will have shifted infinitesimally, of course — you can never step in the same spacetime twice. Nov 20 '19 at 18:31
• @RussellBorogove If you want to go into infinitesimal differences, you could also note that the bolt is being pulled forward by the gravitational attraction from the spacecraft, while the spacecraft is being pulled backward by the gravitational attraction of the bolt. At least one of those forces is negligibly small though - I'm trying to resist the urge to calculate approximate values for them so as to compare them to the gravitational attraction from the planet. Nov 21 '19 at 11:22
• Nitpick: The mass of the craft does not actually cancel out, the force is based on the sum of the masses of the objects. However, in spacecraft orbital mechanics the mass of the craft is so small it's far below the measuring error. Try your equation with the Moon's orbit and you'll have big problems. Nov 22 '19 at 4:43
• @anaximander Real world example--stuff lost from the ISS falls a lot faster than the ISS because it's got a lot less kg per m^2 facing the atmosphere. Nov 22 '19 at 4:44

The acceleration due to gravity will be identical regardless of mass, assuming the mass of your spacecraft is negligible compared to mass of the object you're orbiting. For example the Earths moon is large enough to effect the motion of the earth so it doesn't orbit the centre of the earth, but instead it orbits the shared centre of mass of the Earth and moon (barycenter). Strictly this is true for any orbiting body, but for small objects it is realistic to assume the barycenter is the centre of the earth.

However gravity is not the only force acting on the spacecraft, although it will be strongest up until entry into the earths atmosphere. Drag from the Earths upper atmosphere would probably be noticeable below 2000 km altitude and will accelerate the two objects at different rates causing them to diverge. Also solar radiation pressure will accelerate them at different rates, but this force is so small it would take longer than a single orbit for it to be noticed.

So the two objects would stay at approximately the same distance until the effects of drag in the upper atmosphere start to become measurable.

What the other answers fail to mention is that the mass of your orbiting object actually cancels out. It does not matter. See these two equations:

(1) $$F_1 = F_2 = G m_1 m_2 / r^2$$

(2) $$F_1 = m_1 a_1$$

Where F is force, G is the universal gravitational constant, m is mass, and r is distance between centers of mass of the orbiting and orbited bodies in question. The 1 and 2 represent the object in question, for example $$m_1$$ is the mass of object 1 and $$F_1$$ is the force exerted on object 1.

Thus,

$$a_1 = G m_2 / r^2$$

i.e., the mass of the orbiting object does not influence its acceleration in any way.

edit: added an index of 1 to a.

• Hate to nitpick, but isn't r affected very very very very very very very very very slightly by the change in mass? Nov 20 '19 at 23:45
• r is not a function of mass. It is the mass of the the planet/moon/whatever that is important here (m2). If you want to nitpick, the negligible part I left out was that the moon/planet/whatever accelerates ever-so-slightly toward the orbiting body (m1, in this case). Nov 21 '19 at 2:17
• However, Density (mass distribution) is a function of r. Assuming a spherical planet, discrepancies cancel out though. Nov 21 '19 at 2:24
• That was it,t he center of mass would be every so slightly closer to the planet Nov 22 '19 at 19:32
• Unfortunately, while true, this distracts from the question of orbital speed, which is different from accelaration in radial direction. See the new discussion at space.stackexchange.com/questions/48007/… Oct 15 '20 at 11:48

This is a late answer; a closely related question was recently closed as a duplicate of this.

Does trajectory of an object orbiting a planet depend on the object's mass?

Yes, it does.

Several of the answers correctly invoke the principle of the universality of free fall, which dictates that the acceleration from the perspective of an inertial frame of reference of an object toward the Earth is independent of the object's mass. What these answers miss is that the universality of free fall also dictates that the Earth must be accelerating toward the orbiting object, and this acceleration is directly proportional to the object's mass.

This means that the orbital period of an object orbiting the Earth is $$T = 2\pi\sqrt{\frac{a^3}{G(M+m)}}$$ where $$a$$ is the orbit's semi-major axis length, $$G$$ is the Newtonian gravitational constant, $$M$$ is the mass of the Earth, and $$m$$ is the mass of the orbiting object. This is the Newtonian version of Kepler's third law.

In a theoretical universe in which our Moon was replaced with an Earth-sized object orbiting at 385000 km, that Earth-sized object and the Earth would orbit one another in 19.3 days instead of 27.3 days, the length of a sidereal month. In yet another theoretical universe in which our Moon was replaced with a tiny rock orbiting at 385000 km, the tiny rock would orbit the Earth in 27.5 days instead of 27.3 days.

This is a late answer; a closely related question was recently closed as a duplicate of this.

Does mass of orbiting body affect the orbital speed?

tl;dr: Yes it always does, about half as much. If it's small, like one millionth the mass of the primary, the change in speed is one half of one millionth for example. In the extreme case when the two masses are equal though the trend breaks down and the speed is now 70.7% ($$\sqrt{1/2}$$ ) rather than half.

If you removed the Moon and put a small rock there, it would orbit 0.6% faster than the Moon. Jupiter is about 1/1000 of the Sun or 0.1% the mass. If you removed Jupiter and put a small planet there, it would orbit 0.05% faster than Jupiter does!

Wikipedia's Two body problem and Circular orbit are helpful but I found that the cnx.org page 15. Two body system - circular motion has a particularly straightforward treatment of the circular two body problem. Commons Attribution 4.0 License.

Use

$$r = r_1 + r_2$$

$$m_1 r_1 = m_2 r_2$$

$$\frac{v_1}{r_1} = \frac{v_2}{r_2}$$

$$\omega_1 = \omega_2 = \omega \ \ \text{ orbital angular speed}$$

$$M = m_1 + m_2$$

$$m_2 = M\frac{r_1}{r_1 + r_2}$$

...then some math and physics happens...

$$\omega = \sqrt{\frac{GM}{r^3}} = sqrt{\frac{G(m_1+m_2)}{r^3}}$$

Orbital speed of each body would just be the angular speed $$omega$$ times each body's radius:

$$v_1 = \omega r_1$$

$$v_2 = \omega r_2$$

$$r2 = r \frac{m_1}{M}$$

$$v_2 = \omega r_2 = \omega r \frac{m_1}{M} = \sqrt{\frac{G(m_1+m_2)}{r^3}} r \frac{m_1}{M}$$

It can be shown that if $$m_1$$ (i.e. mass of Earth) is constant and the separation between the two $$r$$ is constant then the change in speed is half as fast as the ratio of masses as long as it's still fairly small.

For example if the mass of the small object is one millionth of the mass of the large object, then the change in speed (compared to massless small object) is one half of one millionth.

For the Moon we have say $$m_2 = m_1 / 81$$, then

$$v_2$$ = 0.9939 $$r_2$$ = 0.9878 $$\omega$$ = 1.0062 and $$\omega r_2$$ = 0.9939

The moon having 1.23% of Earth's mass would move 0.61% slower than a tiny satellite.

This "half the difference" trend breaks down when the two masses become closer to equal.

If the second object were the same mass as the Earth, this trend says the speed would be half of the tiny satellite, but it turns out the speed is $$\sqrt{1/2}$$ or 70.7% rather than 50%. import numpy as np
import matplotlib.pyplot as plt

m1 = 1.0

m2 = np.logspace(-10, 0, 101)

M = m1 + m2

r = 1.0
G = 1

omega = np.sqrt(G * M / r**3)
r2  = r * m1 / M
v2 = omega * r2

plt.figure()
plt.subplot(2, 1, 1)
plt.plot(m2, v2)
plt.xscale('log')
plt.ylim(None, 1.02)
plt.ylabel('v(m2=0) - v "how much slower"')
plt.subplot(2, 1, 2)
plt.plot(m2, 1 - v2)
plt.xscale('log')
plt.yscale('log')
plt.xlabel('m2 with m1 = 1')
plt.ylabel('v(m2=0) - v "how much slower"')
plt.suptitle('G = r = m1 = 1')
plt.show()