Will my mass be greater in deep space compared to when I am on Earth?

Question

Imagine a huge hand picks me up and takes me away from Earth. Will my mass increase as I move away? I am asking this because when we try to separate a nucleus into its constituent nucleons, the total mass of the separated nucleons will be greater than the total mass of the bound nucleus.

Answer 1

Note that in your atomic example, you're not saying that any of the individual nucleons gains mass. Instead we can say that the bound system of nucleons has less mass than the separated systems of nucleons. The mass/energy is gained by the entire system.

The same is true for the gravitational example. The mass of the earth/you system with you on the surface of the earth is (infinitesimally) smaller than the mass of the earth/you system when you are separated by a large distance. But that doesn't mean that we attribute that increase to your mass changing.

Answer 2

Yous say:

when we try to separate a nucleus into its constituent nucleons, the total mass of the separated nucleons will be greater than the total mass of the bound nucleus.

Note that the frame in which you are talking is the quantum mechanical one and the special relativity one, where particles are described by four vectors.

The four-vector of a particle has as a "length" the invariant mass, which characterizes the particle and it is invariant , i.e. in all inertial frames the same. The individual nucleons in the nucleus are what is called "virtual" and the "length" of the four vector describing it is off mass shell, not equal to the invariant mass. So although adding up the quantum numbers (baryon numbers in this case) is the same for the free and the bound nucleons, adding up the masses gives the binding energy of those nucleons in the nucleus.

Thus the analogy cannot hold for the macroscopic case you describe, because the state of "virtual" does not apply to classical gravity, only to quantum mechanical states.

It is true that the four vector describing you on the surface of the earth will be different to the four vector describing you where the hand deposits you, but your invariant mass will be the same due to the four vector algebra,( and the fact that "virtual" is not defined in classical macroscopic gravity).

Answer 3

No matter where something is, its mass is the same. However, if we have a system in a particular configuration (like you and a planet a certain distance apart), and we want to apply forces to the whole system such that the particular configuration remains the same, we must pay for the acceleration of the energy of the configuration as if it was part of the mass of the system.

Here is a concrete, if far-fetched, example involving macroscopic objects with constant mass.

Suppose an indestructible Earth, an indestructible space station in a circular orbit around Earth, and a giant, heavily armed flying saucer in deep space, which is rocketing around in a big circle to stay stationary with respect to both the center of the Earth and the space station. (Like a geosynchronous orbit but instead of staying over one spot on Earth's surface, the flying saucer stays over the space station, continuously expending fuel to stay like that.)

Zim is on the flying saucer. Beth is on the space station. Zim wants to make Beth's accelerometer record an acceleration in a particular direction, but he doesn't want to alter Beth's orbit. So, aiming very carefully, he shoots a continuous stream of countless tiny bullets at Beth's space station and at the same time$^1$ he shoots a continuous stream of countless giant cannon balls at the Earth, so that everyone on Earth will measure an equal change in acceleration. To make it easier to time the simultaneous impacts, Zim uses the same velocity for the bullets and cannon balls.

1: (actually slightly earlier, since the cannon balls must zip past Beth and go a little bit farther to reach the Earth at exactly the same time as the bullets reach Beth)

Zim knows that power equals force times velocity and force equals mass times acceleration, so $P=mav$, and if he wants $a$ and $v$ to be the same for Beth as they are for Earth, he must supply Beth and Earth with the same ratio of power (supplied by the bullets and cannonballs respectively) to their respective masses.

(here $v$ is the velocity of Beth and Earth relative to an inertial frame initially comoving with Beth and the Earth)

• Let $M_e$ be the mass of the earth.

  Let $m_c$ be the mass of a cannon ball.

  Let $m_b$ be the mass of a bullet.

  Let $R$ be the distance between Zim and the Earth's center.

  Let $r_e$ be the radius of the Earth.

  Let $r_b$ be the radius of Beth's orbit.

  Suppose the mass of the space station is negligible compared to the mass of the Earth, such that its gravitational effect on other bodies is zero.

The cannon balls hit the Earth with energy equal to their muzzle energy $T_c = 0.5m_cv_L^2 $ (where $v_L$ is the velocity of the cannon ball relative to Beth and Earth when Zim fires the cannon ball) plus the difference in their gravitational potential energy between their starting position and their ending position $\Delta U_c = Gm_cM_e(1/r_e-1/R)$.

The bullets hit Beth's space station with energy equal to their muzzle energy $T_b = 0.5m_bv_L^2$ plus the difference in their gravitational potential energy between their starting position and their ending position $\Delta U_b = Gm_bM_e(1/r_b-1/R)$.

Then when $r_b$ increases, in order for $$(\Delta U_b + T_b)/m_b $$ to equal $$(\Delta U_c + T_c)/m_c $$ $m_b$ must also increase. That is: the farther from Earth Beth is, the bigger the bullets Zim must use to push her if he doesn't want to change her orbit.