October 2016

Jan Onderco

## Abstract

The purpose of this paper is to analyze gravitational acceleration of a moving inertial reference frame linked to the surface of the Earth in a “momentarily co-moving inertial frame” (MCIF)$^{[1]}$ manner and compare it to the gravitational acceleration analysis of a reference frame that is attached to the gravitational field background. The reason is to investigate how the constant velocity of the inertial reference frame motion, a reference frame on the rotating Earth surface differs from a reference frame on the non-rotating Earth and how the gravitational acceleration affects falling of objects.

A question that needs to be asked is whether a falling object ‘does care’ about any reference frame at all. The intuitive answer is that the gravitational field background is the preferred reference frame.

## Introduction

There is an island on the equator with a $122.5m$ hill, the open ocean to the east and to the west.

The observer shoots projectiles horizontally to the east and to the west at the same time with a velocity $v=465m/s$, we will ignore the air resistance.

What is the time of the falling till the projectiles hit the ocean surface?

The Earth equatorial radius: $6,378.1 km$

The Earth equatorial rotation velocity: $1,674.4 km/h \rightarrow 1674400/3600 = 465m/s$

Figure 1: The variation of both the absolute and the relative values of g with latitude for sea-level conditions.$^{[2]}$

The $g=9.814m/s^{2}$ for non-rotating Earth.

The centrifugal acceleration for $v=465m/s$ is $a_{cf}=v^{2}/r=465^{2}/6378100=0.0339m/s^{2}$.

The rotating Earth gravitational acceleration is $g=9.814-0.0339=9.781m/s^{2}$.

The projectile that was shot to the west is not moving horizontally in the gravitational field therefore it’s going to be falling down with the gravitational acceleration approximately $g=9.814m/s^{2}$.

The projectile that was shot to the east is moving horizontally in the gravitational field with the velocity $v=930m/s$.

The centrifugal acceleration for $v=930m/s$ is $a_{cf}=v^{2}/r=930^{2}/6378100=0.1356m/s^{2}$ therefore it’s going to be falling down with the gravitational acceleration approximately $g=9.814-0.1356=9.678m/s^{2}$.

What are the times of falling? The westward projectile falls $5s$ so $h=0.5gt^{2}=0.5*9.814*25=122.675m$

The eastward projectile falls from the same height in $t=\sqrt{2h/g}=\sqrt{2*122.675/9.678}=5.035s$

The conclusion is that the eastward projectile will be falling $0.035s$ longer than the westward projectile due to the difference in $g$.

## The Inertial Reference Frames

The inertial reference frame at the top of the hill is an approximation only, it is not accurate for the above calculation.

The reference frame that is ‘linked’ to the gravitational field background is preferred one for more accurate calculation.

Figure 2: The gravity field reference frame $K$ (horizontal blue labels) and the Earth reference frame $K’$ (horizontal red labels) at time $T0=0s$.

The horizontal lines are the gravity equipotential lines and the vertical lines are the gravity field lines.

There are two reference frames. The first reference frame $K$, the gravity field frame, has $[0,0]$ at the sea level, straight below the observer that is on the hill at the time $T0=0s$ when the observer shoots the projectiles. This frame is attached to the non-rotating gravitational field. The x-axis values are labeled in blue.

The second reference frame $K’$, the Earth frame, has $[0,0]$ at the sea level, straight below the observer that is on the hill at the time $T0=0s$ when the observer shoots the projectiles but this frame rotates with the Earth at $v=465m/s$. The x-axis values are labeled in red.

The position of the projectiles at $T1=1s$ where fallen height for the blue projectile falling straight down in the gravity field reference frame is $h_{b}=0.5gt^{2}=0.5*9.814*1^{2}=4.907m$ and for the red projectile flying eastward and falling down is $h_{r}=0.5gt^{2}=0.5*9.678*1^{2}=4.839m$.

Figure 3: The gravity field reference frame (horizontal blue labels) and the Earth reference frame (horizontal red labels) at time $T1=1s$.

The position of the projectiles at $T2=2s$ where fallen height for the blue projectile falling straight down in the gravity field reference frame is $h_{b}=0.5gt^{2}=0.5*9.814*2^{2}=19.628m$ and for the red projectile flying eastward and falling down is $h_{r}=0.5gt^{2}=0.5*9.678*2^{2}=19.356m$.

Figure 4: The gravity field reference frame (horizontal blue labels) and the Earth reference frame (horizontal red labels) at time $T2=2s$.

The position of the projectiles at $T3=3s$ where fallen height for the blue projectile falling straight down in the gravity field reference frame is $h_{b}=0.5gt^{2}=0.5*9.814*3^{2}=44.163m$ and for the red projectile flying eastward and falling down is $h_{r}=0.5gt^{2}=0.5*9.678*3^{2}=43.551m$.

The position of the projectiles at $T4=4s$ where fallen height for the blue projectile falling straight down in the gravity field reference frame is $h_{b}=0.5gt^{2}=0.5*9.814*4^{2}=78.512m$ and for the red projectile flying eastward and falling down is $h_{r}=0.5gt^{2}=0.5*9.678*4^{2}=77.424m$.

The position of the projectiles at $T5=5s$ where fallen height for the blue projectile falling straight down in the gravity field reference frame is $h_{b}=0.5gt^{2}=0.5*9.814*5^{2}=122.675m$ and for the red projectile flying eastward and falling down is $h_{r}=0.5gt^{2}=0.5*9.678*5^{2}=120.975m$.

The values are listed in the table below:

time[s] | fallen height of the westward projectile with $g=9.814m/s^{2}$ | fallen height of the eastward projectile with $g=9.678m/s^{2}$ |

1 | $4.907m$ | $4.839m$ |

2 | $19.628m$ | $19.356m$ |

3 | $44.163m$ | $44.163m$ |

4 | $78.512m$ | $77.424m$ |

5 | $122.675m$ | $120.975m$ |

Figure 5: The projectiles fall in the gravity field reference frame. The eastward projectile will be $1.7m$ above the water when the westward projectile hits the ocean surface.

## The International Space Station

There is the International Space Station flying with the velocity $v=7.67km/s$ at about $h=400km$ above the Earth and $g=8.682m/s^{2}$ at that height.

Let us shoot a projectile in the direction of the ISS velocity vector with $v=7.67km/s$ and in the opposite direction of the ISS velocity vector with $v=7.67km/s$.

Now based on what we learned about the gravitational field reference frame we know that the projectile in the direction of the ISS velocity vector will fly away with $2v=15.34km/s$ velocity, $a_{cf}=v^{2}/r=15340^{2}/6778100=34.717m/s^{2}$, the proper centrifugal acceleration needs to be adjusted by subtracting the gravitational acceleration $a_{cf-adj}=34.717-8.682=26.035m/s^{2}$. The projectile in the opposite direction of the ISS velocity vector will be falling straight down because its velocity is $v=0m/s$ in the gravity field reference frame. The ISS ‘inertial reference frame’ comes short in predicting what happens to the projectiles but we are being taught by the General Relativity that there is no force acting on a free falling body.

We are left with a conclusion that a free falling reference frame is an approximation only at some velocities.

## Discussion

It appears the gravitational field reference frame is a preferred one. It provides better prediction of the projectile fall within the gravity field. The question might arise why not to use the reference frame at the center of the gravity field, the Earth center?

The issue is that the locality of the gravitational field determines the gravitational acceleration $g$. We ignored the influence of the Moon in our calculations, but if we want to be accurate we would need to account for the gravitational field coming from the Moon as well, and the Sun, … The projectile fall would be different when the Moon is straight above the shooter, ISS or if it is on the opposite side of the Earth.

## The Curved Space-Time

It appears that the argument mentioned in the previous discussion paragraph is nothing new and it is well know that the effects of the curved space-time are expected to be observed.

Having said that we will consider the following thought experiment. We will mount a very fast 3-axis accelerometer inside of the rifle projectile. The accelerometer has a wireless connection and it can sample and report at pico seconds or faster. We want to measure an instantaneous acceleration when $dt \rightarrow 0$ in order to make the curved space-time as flat as possible.

Figure 6: The rifles with fast accelerometers inside of the projectiles moving at constant velocity at the end of the barrels.

The projectiles have the instantaneous velocities in the gravitational field reference frame shown in blue; $0m/s$ for the projectile moving westward and $930m/s$ for the projectile moving eastward. The velocity difference will affect the instantaneous acceleration as well. The westward accelerometer will report the gravitational acceleration of $9.814m/^{2}$ and the eastward accelerometer will report the gravitational acceleration of $9.678m/^{2}$.

## Discussion #2

A point on the Earth surface at the equator moves $0.465nm$ in one picosecond in the gravitational field reference frame. The curved space-time can be considered flat in one picosecond and the traveled distance can be considered as a constant straight line motion, still it can be detected by the thought experiment.

## The International Space Station and the super rifle

The extended thought experiment places a super rifle, capable to shoot its projectile at 7.67km/s, inside the International Space Station firmly horizontally attached to it with the $y$ axis going through the center of the Earth.

Figure 7: The super rifle with fast accelerometers inside of the projectile moving at constant velocity at the end of the barrel inside the IIS. The blue reference frame is linked to the gravitational field background. The red reference frame is linked to the ISS.

The ISS moves at $7.67km/s$ in the blue reference frame that is linked to the gravitational field background. The red reference frame is linked to the ISS. The projectile is being shot in the opposite direction of the ISS motion velocity vector. The projectile reaches constant velocity at the end of the barrel that is $7.67km/s$ in the red ISS reference frame and it is $0km/s$ in the blue gravitational field background reference frame. At this moment the accelerometer inside the projectile with a wireless connection and a capability to measure in femto seconds is motion less in the blue gravitational field reference frame and it is still inside of the barrel having the support from beneath therefore it is not in a free fall and the accelerometer measures the gravitational acceleration $8.682m/s^2$.

When the projectile exits the barrel then it loses the support from beneath and it enters into the free fall. The accelerometer reports $0m/s^2$ gravitational acceleration.

Figure 8: The super rifle with fast accelerometer inside of the projectile moving at constant velocity outside of the barrel inside the IIS.

## Discussion #3

The blue gravitational field reference frame predicts properly the physics. It predicts accurately the accelerometer readings.

The red International Space Station free falling reference frame does not predict the physics accurately. It does not predict the change in the accelerometer readings between the projectile position inside of the barrel and outside of the barrel.

## The International Space Station and the super rifle in the intergalactic space

Figure 9: The super rifle with fast accelerometers inside of the projectile moving at constant velocity at the end of the barrel inside the IIS. The ISS is in the intergalactic space where $g$ is close to $0$.

The accelerometer inside of the projectile can not measure any gravitational acceleration even when the projectile is inside the barrel because there is almost none in the intergalactic space.

## Conclusion

We can see that the physics in the free falling ISS in the intergalactic space are different from the physics in the free falling ISS in the Earth gravitational field. The physics of the free falling reference frames are dependent on the motion through the gravitational field and therefore the free falling reference frames are not equivalent.

## References

${[}1{]}$ Does a clock’s acceleration affect its timing rate?

${[}2{]}$ From Engineering Mechanics, Volume 2, Dynamics, Seventh Edition by J. L. Meriam, L. G. Kraige, Chapter 1, page 10