Why do spacecraft always experience a black-out area that disrupts communications when they return to Earth?

Abstract

In this paper, Lorentz factor and Lorentz transformations are modified based on the new aether theory, and the blackout which leads to communication interruption of high-speed moving objects is analyzed by using the modified Lorentz electromagnetic field transformation formula. It is concluded that blackout is caused by the change of wave form of electromagnetic wave from stationary energy space (etheric reference frame) to moving medium (etheric reference frame) or from moving medium to stationary energy space.

Main article text

INTRODUCTION

When spacecraft re-enter the atmosphere and return to the earth at a high speed, communication with the ground will be completely interrupted at a certain height, this is the blackout [1]. The blackout appears between 35 km and 80 km above the earth. How is the blackout formed? The current scientific consensus is that all spacecraft re-enter the atmosphere at such high speeds that the friction with the atmosphere causes the surface of the aircraft to be so hot that it ionizes both the gas and the ablated heat-resistant material. As a result, a sheath of hot plasma forms around the spacecraft, which causes radio waves to decay or reflect, disrupting radio communication between the ground and the spacecraft. Since the 1950s, people began to study the blackout and its elimination. However, until today, blackout is still a difficult problem to the scientific community. When Mars landers land on Mars, they will also generate blackout that disrupt communications [2–4]. Because the Martian atmosphere is very thin, its density is less than 1% of the Earth atmosphere, and the initial speed of the Mars lander entering the Martian atmosphere to generate the blackout is between 4.7 km/s and 7.26 km/s [2], which is lower than the initial speed of the Earth return capsule entering the blackout region (greater than the first cosmic speed of 7.9 km/s). So the idea that the blackout created by the Mars lander is caused by a sheath of plasma created by the lander’s rapid friction with the Martian atmosphere does not stand up to scrutiny. The purpose of this study is to find out the real cause of the blackout.

The modifications of Lorentz factor and Lorentz transformations

In Section 2.1 of reference [5], The Physical Mechanism of Constancy of Light Velocity, the author puts forward a new aether theory, which is as follows:

1. The cosmic space is full of aether (background energy), and aether is also the medium in which light waves travel.

2. All fields (gravitational field, electromagnetic field, strong interaction field and weak interaction field etc.) contain and compress aether (energy).

3. Each particle or object carries an aetheric layer (sphere of influence) that surrounds and moves with it.

It should be emphasized that the new aether theory believes that energy is the most basic substance that constitutes the universe (all matter and space-time), and this most basic substance is the aether. According to this new aether theory, the speed effects such as the ruler shrinking, the clock slowing down and the mass increasing are essentially aether (energy) effects, that is, the physical effects of the aether acting on the object when the object is moving in the aether, and these physical effects are objective and independent of the observer’s choice of reference frame. Since the energy (aether) density of space is different everywhere in the universe, and the strength of aether effect produced by objects moving at the same speed in space with different energy (aether) density should be different. Therefore, the Lorentz factor must be modified to reflect the effect of spatial energy (aether) density:

(1) $${\gamma }^{\prime }=\frac{1}{\sqrt{1-{{\beta }^{\prime }}^2}}$$

(2) $${\beta }^{\prime }=\text{f}(\rho )\text{v/c}$$

In formula (2), f(ρ) is a function of ρ, which is a dimensionless number.

(3) $$\text{f}(\rho )=\text{k}{\rho }^{\alpha }$$

In formula (3), α and k are constants, ρ is the energy density of the space in which the moving object resides. The values of α and k are determined by experiment.

In all Lorentz transformations (Lorentz coordinate transformations, velocity transformations, electromagnetic field transformations), change the speed v to f(ρ)v, and change the Lorentz factor γ to the modified Lorentz factor γ′, we obtain the modified Lorentz transformations:

1. Modified Lorentz coordinate transformations:

   (4) $$\{\begin{array}{llll}\text{x}\hfill & =\hfill & {\gamma }^{\prime }[\text{x}\prime +\text{f}(\rho )\text{vt}\prime ]\hfill & (4.1)\hfill \\ \text{y}\hfill & =\hfill & \text{y}\prime \hfill & (4.2)\hfill \\ \text{z}\hfill & =\hfill & \text{z}\prime \hfill & (4.3)\hfill \\ \text{t}\hfill & =\hfill & {\gamma }^{\prime }[\text{t}+\text{f}(\rho )\text{vx}\prime /{\text{c}}^2]\hfill & (4.4)\hfill \end{array}$$

   and the modified Lorentz coordinate inverse transformations:

   (5) $$\{\begin{array}{llll}\text{x}\prime \hfill & =\hfill & {\gamma }^{\prime }[\text{x}-\text{f}(\rho )\text{vt}]\hfill & (5.1)\hfill \\ \text{y}\prime \hfill & =\hfill & \text{y}\hfill & (5.2)\hfill \\ \text{z}\prime \hfill & =\hfill & \text{z}\hfill & (5.3)\hfill \\ \text{t}\prime \hfill & =\hfill & {\gamma }^{\prime }[\text{t}-\text{f}(\rho ){\text{vx/c}}^2]\hfill & (5.4)\hfill \end{array}$$

2. Modified Lorentz velocity transformations:

   (6) $$\{\begin{array}{llll}{\text{u}}_{\text{x}}\hfill & =\hfill & \frac{\text{u}{\prime }_{\text{x}}+\text{f}(\rho )\text{v}}{1+\text{f}(\rho )\text{vu}{\prime }_{\text{x}}{\text{/c}}^2}\hfill & (6.1)\hfill \\ {\text{u}}_{\text{y}}\hfill & =\hfill & \frac{\text{u}{\prime }_{\text{y}}}{{\gamma }^{\prime }[1+\text{f}(\rho )\text{vu}{\prime }_{\text{x}}{\text{/c}}^2]}\hfill & (6.2)\hfill \\ {\text{u}}_{\text{z}}\hfill & =\hfill & \frac{\text{u}{\prime }_{\text{z}}}{{\gamma }^{\prime }[1+\text{f}(\rho ){\text{vu}}_{\text{x}}^{\prime }{\text{/c}}^2]}\hfill & (6.3)\hfill \end{array}$$

   and the modified Lorentz velocity inverse transformations:

   (7) $$\{\begin{array}{llll}\text{u}{\prime }_{\text{x}}\hfill & =\hfill & \frac{{\text{u}}_{\text{x}}-\text{f}(\rho )\text{v}}{1-\text{f}(\rho ){\text{vu}}_{\text{x}}{\text{/c}}^2}\hfill & (7.1)\hfill \\ \text{u}{\prime }_{\text{y}}\hfill & =\hfill & \frac{{\text{u}}_{\text{y}}}{{\gamma }^{\prime }[1-\text{f}(\rho ){\text{vu}}_{\text{x}}{\text{/c}}^2]}\hfill & (7.2)\hfill \\ \text{u}{\prime }_{\text{z}}\hfill & =\hfill & \frac{{\text{u}}_{\text{z}}}{{\gamma }^{\prime }[1-\text{f}(\rho ){\text{vu}}_{\text{x}}{\text{/c}}^2]}\hfill & (7.3)\hfill \end{array}$$

3. Modified Lorentz electromagnetic field transformations:

   (8) $$\{\begin{array}{llll}\text{E}{\prime }_{\text{x}}\hfill & =\hfill & {\text{E}}_{\text{x}}\hfill & (8.1)\hfill \\ \text{E}{\prime }_{\text{y}}\hfill & =\hfill & {\gamma }^{\prime }[{\text{E}}_{\text{y}}-\text{f}(\rho ){\text{vB}}_{\text{z}}]\hfill & (8.2)\hfill \\ \text{E}{\prime }_{\text{z}}\hfill & =\hfill & {\gamma }^{\prime }[{\text{E}}_{\text{z}}+\text{f}(\rho ){\text{vB}}_{\text{y}}]\hfill & (8.3)\hfill \\ \text{B}{\prime }_{\text{x}}\hfill & =\hfill & {\text{B}}_{\text{x}}\hfill & (8.4)\hfill \\ \text{B}{\prime }_{\text{y}}\hfill & =\hfill & {\gamma }^{\prime }\left[{\text{B}}_{\text{y}}+\frac{\text{f}(\rho )\text{v}}{{\text{c}}^2}{\text{E}}_{\text{z}}\right]\hfill & (8.5)\hfill \\ \text{B}{\prime }_{\text{z}}\hfill & =\hfill & {\gamma }^{\prime }\left[{\text{B}}_{\text{z}}-\frac{\text{f}(\rho )\text{v}}{{\text{c}}^2}{\text{E}}_{\text{y}}\right]\hfill & (8.6)\hfill \end{array}$$

   and Modified Lorentz electromagnetic field inverse transformations:

   (9) $$\{\begin{array}{llll}{\text{E}}_{\text{x}}\hfill & =\hfill & \text{E}{\prime }_{\text{x}}\hfill & (9.1)\hfill \\ {\text{E}}_{\text{y}}\hfill & =\hfill & {\gamma }^{\prime }[\text{E}{\prime }_{\text{y}}+\text{f}(\rho )\text{vB}{\prime }_{\text{z}}]\hfill & (9.2)\hfill \\ {\text{E}}_{\text{z}}\hfill & =\hfill & {\gamma }^{\prime }[\text{E}{\prime }_{\text{z}}-\text{f}(\rho )\text{vB}{\prime }_{\text{y}}]\hfill & (9.3)\hfill \\ {\text{B}}_{\text{x}}\hfill & =\hfill & \text{B}{\prime }_{\text{x}}\hfill & (9.4)\hfill \\ {\text{B}}_{\text{y}}\hfill & =\hfill & {\gamma }^{\prime }\left[\text{B}{\prime }_{\text{y}}-\frac{\text{f}(\rho )\text{v}}{{\text{c}}^2}\text{E}{\prime }_{\text{z}}\right]\hfill & (9.5)\hfill \\ {\text{B}}_{\text{z}}\hfill & =\hfill & {\gamma }^{\prime }\left[\text{B}{\prime }_{\text{z}}+\frac{\text{f}(\rho )\text{v}}{{\text{c}}^2}\text{E}{\prime }_{\text{y}}\right]\hfill & (9.6)\hfill \end{array}$$

In fact, the modified Lorentz factor and the modified Lorentz Transformations can be derived from the modified relativity principle and the modified invariant speed of light principle. For details, please see reference [5], P1623-P1626, 4. on the Lorentz Transformation. The modified Lorentz factor (formula (1)) and the modified Lorentz transformation formula (formula (4)–(7)) in this paper can be obtained by replacing the velocity v with f(ρ)v in all derivations. In the derivation process of the electromagnetic field transformation formula (Ref. [6]), it is only necessary to replace v with f(ρ)v to obtain the modified electromagnetic field transformations in this paper (formulas (8.1)–(8.6) and (9.1)–(9.6)).

In-depth analysis of the cause of the blackout

Due to the earth’s gravitational field, there exists a relatively static etheric (space energy) layer on the earth’s surface [5]. Therefore, on the Earth, the space layer at a certain height on the earth’s surface can be regarded as a relatively static medium. In the earth etheric layer, electromagnetic waves emanating from antennas at rest relative to the earth etheric layer obey maxwell’s equations of the stationary medium. Because the etheric space formed by the superposition of all nuclear etheric layers inside the antenna (wire) is static relative to the wire, it can be regarded as a medium that is completely static relative to the wire, therefore, the electromagnetic waves generated by the free electron motion in the wire obey the Maxwell equations of the stationary medium inside the wire and in the very thin space layer on the surface of the wire. This study involves the antenna of the return capsule, the Earth etheric layer and the ground antenna. In the case of a spacecraft reentry capsule returning to Earth at a high speed, what happens when the capsule receives the radio signal from the ground antenna, and when the ground antenna receives the radio signal from the capsule antenna? Let’s analyze separately.

1. When the antenna of the capsule receives the radio signal from the ground antenna

   Since the ground antenna is stationary relative to the ground, the ground antenna can be regarded as the earth aether reference system. In this case, the Earth etheric layer can be regarded as the stationary medium relative to the earth, and the antenna of the re-entry capsule can be regarded as the medium moving relative to the Earth etheric layer. Fix coordinate system oxyz on reference frame K (Earth) and coordinate system o′x′y′z′ on reference frame K′ (re-entry module antenna). For convenience, assume that the corresponding coordinate axes of the two coordinate systems are parallel to each other, and assume that K′ moves in the positive direction of the x axis with respect to K at velocity V, and that the origin o and o′ of the two coordinate systems coincide when t=t′. According to the modified relativity principle [5], the physical theorems are of the same form in all etheric systems. Therefore, when observed in the K system (the earth aether layer) at rest, the electromagnetic wave emitted by the ground antenna satisfies Maxwell’s equations of stationary medium:

   (10) $$\left\{\begin{array}{llll}\nabla \hfill & \cdot \hfill & \text{E}=\frac{\rho }{{\epsilon }_0}\hfill & (10.1)\hfill \\ \nabla \hfill & \cdot \hfill & \text{B}=0\hfill & (10.2)\hfill \\ \nabla \hfill & \times \hfill & \text{E}=-\frac{\partial \text{B}}{\partial \text{t}}\hfill & (10.3)\hfill \\ \nabla \hfill & \times \hfill & \text{B}={\mu }_0\text{J}+{\epsilon }_0{\mu }_0\frac{\partial \text{E}}{\partial \text{t}}\hfill & (10.4)\hfill \end{array}\right.$$

   According to the modified Lorentz transformations of electromagnetic field (8.1)–(8.6), we can get the electromagnetic field components measured in K′ system (re-entry capsule antenna) as follows:

   (11) $$\{\begin{array}{llll}\text{E}{\prime }_{\text{x}}\hfill & =\hfill & {\text{E}}_{\text{x}}\hfill & (11.1)\hfill \\ \text{E}{\prime }_{\text{y}}\hfill & =\hfill & {\gamma }^{\prime }[{\text{E}}_{\text{y}}-\text{f}(\rho ){\text{vB}}_{\text{z}}]\hfill & (11.2)\hfill \\ \text{E}{\prime }_{\text{z}}\hfill & =\hfill & {\gamma }^{\prime }[{\text{E}}_{\text{z}}+\text{f}(\rho ){\text{vB}}_{\text{y}}]\hfill & (11.3)\hfill \\ \text{B}{\prime }_{\text{x}}\hfill & =\hfill & {\text{B}}_{\text{x}}\hfill & (11.4)\hfill \\ \text{B}{\prime }_{\text{y}}\hfill & =\hfill & {\gamma }^{\prime }\left[{\text{B}}_{\text{y}}+\frac{\text{f}(\rho )\text{v}}{{\text{c}}^2}{\text{E}}_{\text{z}}\right]\hfill & (11.5)\hfill \\ \text{B}{\prime }_{\text{z}}\hfill & =\hfill & {\gamma }^{\prime }\left[{\text{B}}_{\text{z}}-\frac{\text{f}(\rho )\text{v}}{{\text{c}}^2}{\text{E}}_{\text{y}}\right]\hfill & (11.6)\hfill \end{array}$$

   In the above formula, ρ is the energy (etheric) density of the space in which the re-entry capsule resides. From equations (11.2), (11.3), (11.5), (11.6), (1) and (2), it can be seen that when f(ρ)v→C, we can get γ′→∞. Therefore, the higher the f(ρ)v is, the greater the deformation of electromagnetic wave received by the antenna of the re- entry capsule will be (i.e., the greater the difference between ${{\text{E}}^{\prime }}_{\text{y}}$ and E_y, ${{\text{E}}^{\prime }}_{\text{z}}$ and E_z, ${{\text{B}}^{\prime }}_{\text{y}}$ and B_y, ${{\text{B}}^{\prime }}_{\text{z}}$ and B_z, $\text{E}{\prime }_{\text{y}}$ and ${{\text{E}}^{\prime }}_{\text{z}},{\text{B}\mathrm{\hspace{0.17em}}\prime }_{\text{y}}$ and $\text{B}{\prime }_{\text{z}}$ ), thus resulting in a lower signal-to-noise ratio of the signal, so the signal received By the antenna of the re-entry capsule will be worse. In addition to the size of f(ρ)v, the speed of change of f(ρ)v also contributes to the formation of the blackout. From the formulas (11.2), (11.3), (11.5) and (11.6), it can be seen that the faster the change of f(ρ)v, the faster the change of the size of γ′, $\text{E}{\prime }_{\text{y}},\text{E}{\prime }_{\text{z}},\text{B}{\prime }_{\text{y}}$ and $\text{B}{\prime }_{\text{z}},$ so the greater the impact on the signal quality of communication.

2. When the ground antenna receives the radio signal from the antenna of the capsule

   Since the ground antenna is stationary relative to the ground, the ground antenna can be regarded as the earth aether reference system. Fix coordinate system oxyz on reference frame K (Earth) and coordinate system o′x′y′z′ on reference frame K′ (re-entry module antenna). For convenience, assume that the corresponding coordinate axes of the two coordinate systems are parallel to each other, and assume that K′ moves in the positive direction of the x axis with respect to K at velocity v, and that the origin o and o′ of the two coordinate systems coincide when t=t′. Therefore, when observed in the K′ system (re-entry module antenna) at rest, the electromagnetic wave emitted by the re-entry module antenna satisfies Maxwell’s equations of stationary medium:

   (12) $$\{\begin{array}{llll}\nabla \hfill & \cdot \hfill & \text{E}\prime =\frac{{\rho }^{\prime }}{{\epsilon }_0}\hfill & (12.1)\hfill \\ \nabla \hfill & \cdot \hfill & \text{B}\prime =0\hfill & (12.2)\hfill \\ \nabla \hfill & \times \hfill & \text{E}\prime =-\frac{\partial \text{B}\prime }{\partial \text{t}}\hfill & (12.3)\hfill \\ \nabla \hfill & \times \hfill & \text{B}\prime ={\mu }_0\text{J}\prime +{\epsilon }_0{\mu }_0\frac{\partial \text{E}\prime }{\partial \text{t}}\hfill & (12.4)\hfill \end{array}$$

   According to the modified Lorentz transformations of electromagnetic field (9.1)–(9.6), we can get the electromagnetic field components measured in K system (the earth aether layer) as follows:

   (13) $$\{\begin{array}{llll}{\text{E}}_{\text{x}}\hfill & =\hfill & \text{E}{\prime }_{\text{x}}\hfill & (13.1)\hfill \\ {\text{E}}_{\text{y}}\hfill & =\hfill & {\gamma }^{\prime }[\text{E}{\prime }_{\text{y}}+\text{f}(\rho )\text{vB}{\prime }_{\text{z}}]\hfill & (13.2)\hfill \\ {\text{E}}_{\text{z}}\hfill & =\hfill & {\gamma }^{\prime }[\text{E}{\prime }_{\text{z}}-\text{f}(\rho )\text{vB}{\prime }_{\text{y}}]\hfill & (13.3)\hfill \\ {\text{B}}_{\text{x}}\hfill & =\hfill & \text{B}{\prime }_{\text{x}}\hfill & (13.4)\hfill \\ {\text{B}}_{\text{y}}\hfill & =\hfill & {\gamma }^{\prime }\left[\text{B}{\prime }_{\text{y}}-\frac{\text{f}(\rho )\text{v}}{{\text{c}}^2}\text{E}{\prime }_{\text{z}}\right]\hfill & (13.5)\hfill \\ {\text{B}}_{\text{z}}\hfill & =\hfill & {\gamma }^{\prime }\left[\text{B}{\prime }_{\text{z}}+\frac{\text{f}(\rho )\text{v}}{{\text{c}}^2}\text{E}{\prime }_{\text{y}}\right]\hfill & (13.6)\hfill \end{array}$$

In the above formula, ρ is the energy (etheric) density of the space in which the re-entry capsule resides. From equations (13.2), (13.3), (13.5), (13.6), (1) and (2), it can be seen that when f(ρ)v→C, we get γ′→∞. Therefore, the higher the f(ρ)v is, the greater the deformation of the electromagnetic wave received by the ground antenna is (i.e., the greater the difference between Ey and Ey′, Ez and Ez′, By and By′, Bz and Bz′, Ey and Ez, By and Bz), thus resulting in the lower the signal-to-noise ratio of the signal received by the ground antenna, and the worse the signal received by the ground antenna. In addition to the size of f(ρ)v, the speed of change of f(ρ)v also contributes to the formation of the blackout. From the formulas (13.2), (13.3), (13.5) and (13.6), it can be seen that the faster the change of f(ρ)v, the faster the change of the size of γ′, E_y, E_z, B_y and B_z, so the greater the impact on the signal quality of communication.

Therefore, when the re-entry capsule quickly enters the Earth aether layer, the signal-to- noise ratio of the radio signals received by the antenna of the re-entry capsule from the ground antenna and the radio signals received by the ground antenna from the antenna of the re-entry capsule will be very low, resulting in communication interruption. After the re-entry capsule enters the atmosphere, due to the resistance of the atmosphere, the speed of the re-entry capsule will be constantly reduced. Since f(ρ) will increase with the decrease of altitude, the re-entry capsule will not immediately get out of the blackout- area even though the speed of the re-entry capsule decreases with the decrease of altitude under the resistance of air. When the velocity of the re-entry capsule is less than a certain value, the SNR of the radio signals received by the antenna of the re-entry capsule and the ground antenna will be greater than a certain threshold, so that the antenna of the re-entry capsule can receive the radio signals from the ground antenna, and the ground antenna can receive the radio signals sent by the antenna of the re-entry capsule. That’s why the re-entry capsule experiences a blackout- area when it returns to Earth.

The specific process of producing the blackout when the re-entry capsule returns to the Earth is as follows: when the re-entry capsule just enters the atmosphere, the gravitational field strength is not strong enough, that is, f(ρ) is not large enough, so γ′ is not large enough, so there is no blackout. As the height of the capsule continues to decrease, the velocity of the capsule and the strength of the gravitational field are increasing, that is, f(ρ)v (or γ′) is increasing, and when f(ρ)v (or γ′) is greater than a certain value, the blackout begins to emerge. On the other hand, as the altitude of the re-entry capsule continues to decrease, the atmospheric density of the atmosphere also continues to increase, which causes the speed of the re-entry capsule to increase to a certain maximum and then start to decline. When the height drops to a certain degree, f(ρ)v (or γ′) begins to decrease, and when f(ρ)v (or γ′) is small to a certain value, the blackout disappears.

Due to the gravity of the planet, the surface of any planet (Mars, Jupiter, moon, etc.) has the etheric (space energy) layer which is stationary relative to the planet and moves with the planet. Therefore, the space layer at a certain height on the surface of the planet can be regarded as the stationary medium relative to the planet. So we can predict that when a spacecraft enters the aether (space energy) layer of any planet, whenever it is fast enough, it will experience a blackout-area that disrupts radio communication.

Further modification of Lorentz factor and Lorentz transformations

Considering that when an object is stationary in space, if the strength of the gravitational field of space is increasing, there will still be time dilation, mass increase and space contraction effects, so it is necessary to further modify the Lorentz factor. If the spacecraft is out of the gravitational clutches of all the planets and stars and is in the galaxy and orbiting the center of the galaxy, then we have [7,8]:

(14) $$\text{F}=\frac{{\text{mu}}^2}{\text{r}}={\text{G}}_0\text{m}{\displaystyle \sum _{\text{i}=1}^{\text{n}}\frac{{\text{M}}_{\text{i}}}{{\text{r}}_{\text{i}}^2}}$$

In the above formula, m is the mass of the spacecraft, u is the velocity of the spacecraft around the center of the galaxy, r is the distance from the spacecraft to the center of the galaxy, G₀ is the constant, M_i is the mass of any object in the galaxy, r_i is the distance from the center of mass of M_i to the center of mass of m, and n is the total number of objects in the galaxy. This leads to:

(15) $${\text{u}}^2={\text{G}}_0\text{r}{\displaystyle \sum _{\text{i}=1}^{\text{n}}\frac{{\text{M}}_{\text{i}}}{{\text{r}}_{\text{i}}^2}}$$

(16) $${\gamma }^{″}=\frac{1}{\sqrt{1-\frac{u^2}{{\text{c}}^2}}}=\frac{1}{\sqrt{1-\frac{{\text{G}}_0\text{r}}{{\text{c}}^2}{\displaystyle {\sum }_{\text{i}=1}^{\text{n}}\frac{{\text{M}}_{\text{i}}}{{\text{r}}_{\text{i}}^2}}}}$$

(17) $${{\gamma }^{\prime }}^{″}={\gamma }^{\prime }\times {{\gamma }^{\prime }}^{\prime }=\frac{1}{\sqrt{\left(1-\frac{\text{f}{(\rho )}^2{\text{v}}^2}{{\text{c}}^2}\right)\times \left(1-\frac{{\text{G}}_0\text{r}}{{\text{c}}^2}{\displaystyle {\sum }_{\text{i}=1}^{\text{n}}\frac{{\text{M}}_{\text{i}}}{{\text{r}}_{\text{i}}^{\text{2}}}}\right)}}$$

When a spacecraft approaches any of the planets [8,9]:

(18) $$\text{F}=\frac{{\text{mu}}^{\text{2}}}{\text{r}}={\text{G}}_0\text{m}{\displaystyle \sum _{\text{i}=1}^{\text{n}}\frac{{\text{M}}_{\text{i}}}{{\text{r}}_{\text{i}}^2}}\approx \frac{\text{GmM}}{{\text{r}}^2}$$

In the above equation, m is the mass of the spacecraft, u is the velocity of the spacecraft around the planet, r is the distance of the spacecraft to the center of the planet, and G is the gravitational constant. This leads to:

(19) $${\text{u}}^2=\frac{\text{GM}}{\text{r}}$$

(20) $${\gamma }^{″}=\frac{1}{\sqrt{1-\frac{{\text{u}}^2}{{\text{c}}^2}}}=\frac{1}{\sqrt{1-\frac{\text{GM}}{{\text{c}}^2\text{r}}}}$$

(21) $${\gamma }^{‴}={\gamma }^{\prime }\times {\gamma }^{″}=\frac{1}{\sqrt{\left(1-\frac{\text{f}{(\rho )}^2{\text{v}}^2}{{\text{c}}^2}\right)\times \left(1-\frac{\text{GM}}{{\text{c}}^2\text{r}}\right)}}$$

Change the Lorentz factor γ to the modified Lorentz factor γ″′ in all Lorentz transformations and we get the new modified Lorentz transformations. In particular, the following new formulas for time dilation, mass increase, and orbital radius contraction of orbital electrons are obtained (see formula (5.31) in reference [5]):

(22) $$\text{t}=\frac{{\text{t}}_0}{{\gamma }^{‴}}={\text{t}}_0\sqrt{\left(1-\frac{\text{f}{(\rho )}^2{\text{v}}^2}{{\text{c}}^2}\right)\times \left(1-\frac{\text{GM}}{{\text{c}}^2\text{r}}\right)}$$

(23) $$\text{m}={\text{m}}_{\text{0}}{\gamma }^{‴}=\frac{{\text{m}}_0}{\sqrt{\left(1-\frac{\text{f}{(\rho )}^2{\text{v}}^2}{{\text{c}}^2}\right)\times \left(1-\frac{\text{GM}}{{\text{c}}^2\text{r}}\right)}}$$

(24) $${\text{r}}_{\text{a}}=\frac{{\text{r}}_{\text{a}0}}{{\gamma }^{‴}}={\text{r}}_{\text{a}0}\sqrt{\left(1-\frac{\text{f}{(\rho )}^2{\text{v}}^2}{{\text{c}}^2}\right)\times \left(1-\frac{\text{GM}}{{\text{c}}^2\text{r}}\right)}$$

When v=0:

(25) $${\gamma }^{‴}={\gamma }^{″}=\frac{1}{\sqrt{1-\frac{\text{GM}}{{\text{c}}^2\text{r}}}}$$

(26) $$\text{t}=\frac{{\text{t}}_0}{{\gamma }^{″}}={\text{t}}_0\sqrt{1-\frac{\text{GM}}{{\text{c}}^2\text{r}}}$$

(27) $$\text{m}={\text{m}}_{\text{0}}{\gamma }^{″}=\frac{{\text{m}}_0}{\sqrt{1-\frac{\text{GM}}{{\text{c}}^2\text{r}}}}$$

(28) $${\text{r}}_{\text{a}}=\frac{{\text{r}}_{\text{a}0}}{{\gamma }^{″}}={\text{r}}_{\text{a}0}\sqrt{1-\frac{\text{GM}}{{\text{c}}^2\text{r}}}$$

CONCLUSIONS

The Lorentz factor is related not only to velocity but also to space energy density. When the radio signal enters the moving ether (energy) space from the static ether (energy) space, if the Lorentz factor is greater than a certain value, the communication blackout will be generated, and if the Lorentz factor is changing, the value of the Lorentz factor generated by the communication blackout will be reduced. If the mass of the planet is large enough, a spacecraft landing on the planet will create a blackout due to the large Lorentz factor and the rapid change of the Lorentz factor.

DISCUSSIONS

1. Is bad communication on the high-speed train caused by blackout?

   Consider that the reentry module entered the blackout range at an altitude of 80 km at a speed of 7,900 m/s and exited the blackout range at an altitude of 35 km at a speed of 200 m/s, although f(ρ)v was reduced when exiting the blackout range (this may reduce the speed requirement for generating the blackout). However, because the closer to the Earth’s surface, the stronger the gravitational field strength, that is, the larger the f(ρ), It can be inferred that an object moving close to the ground produces a blackout at a lower speed than an object moving at an altitude of 35 km. So it is reasonable to guess that the reason for the poor communication signal of trains at speeds greater than 200 km/h (about 56 m/s) is the same reason that the blackout occurs when the spacecraft reentry module returns to Earth.

2. Why did the spacecraft not create a blackout during the landing on the moon?

   As a spacecraft orbits the moon, its centripetal force is provided by the moon’s gravitational pull on the spacecraft. According to the formula of centripetal force and gravitation, it can be obtained:

   (29) $$\text{v}=\sqrt{\frac{\text{GM}}{\text{r}}}$$

   In the above formula, G is the gravitational constant, M is the mass of the moon, and r is the radius of the spacecraft’s orbit around the moon. Assuming an orbital altitude of about 100 km, since the radius of the moon is about 1737.4 km, then the orbital radius of the spacecraft r=1837.4 km. The mass of the moon is M=7.349×10²² kg, and the gravitational constant G=6.67×10^-11 N·m²/kg². By bringing the above data into the formula (29), we get: v≈1.68 km/s. Because the initial velocity of the spacecraft landing on the moon (1.68 km/s) is far less than the initial velocity of the spacecraft landing on Mars entering the blackout (4.7 km/s ~ 7.26 km/s) [2], and the gravitational field intensity of the lunar surface is only about 43% of the surface of Mars, so the f(ρ)v value of the spacecraft landing on the moon is not large enough, so no blackout is generated.

3. Why didn’t unmanned spacecraft like Voyager and Pioneer experience blackouts during their flight to the edge of the solar system and beyond?

Because Voyager and Pioneer did not land on any planets during their flight to the edge of the solar system, the gravitational field strength (energy density) of space was not high enough, that is, the f(ρ)v value was not large enough, so there was no blackout phenomenon.

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