Moonshots just hit its name by headlining a mission to Alpha Centauri.
So:
It takes light approximately 4.24 to 4.37 years to travel from Earth to the Alpha Centauri star system.
Now we go complicated:
For a rocket accelerating at 1G (\(9.81\text{ m/s}^2\)) for the first half of the journey and decelerating at 1G for the second half, the travel time depends entirely on who is measuring it due to the effects of special relativity.
Because the rocket reaches a top speed of roughly 95% the speed of light 0.95c at the midpoint, time dilation becomes significant:
From the Crew's Perspective (Proper Time): The journey takes about 3.5 to 3.6 years.
From Earth's Perspective (Coordinate Time): The journey takes about 5.9 to 6.0 years.
Space X is out:
Pure Photon/Antimatter Drive (100% efficient) 3,700 tons
Theoretical Maximum. Requires equal amounts of matter and antimatter converting perfectly into light.
Advanced Nuclear Fusion (Efficiency ~1%,)10¹⁶ tons
Impossible. This requires a ball of hydrogen fuel the size of a small moon just to move a 100-ton ship.
Chemical Rockets (Today's tech) More than the atoms in the universe
Physically Impossible. The exponential curve of the rocket equation breaks completely.
Or in different words:
This is roughly 1,000 times the total global energy reserves available on Earth.
Not much more realistic:
A standard human satellite or deep-space probe would take between 70,000 and 75,000 years to reach Alpha Centauri.
If you used humanity's absolute fastest uncrewed probe—the Parker Solar Probe, which relies on the Sun's immense gravity to whip it up to 692,000 km/h (430,000 mph)—the trip would still take about 6,700 to 6,800 years.
The most distant human-made object that NASA still actively communicates with is Voyager 1
Where it stands compared to Alpha Centauri: Despite traveling since 1977, Voyager 1 has only covered about 0.00027 light-years. Alpha Centauri is 4.37 light-years away—making the neighbor star system over 16,000 times farther out than where Voyager 1 is right now.
With all given respect to the collective presence of PhDs, I would not have invited the guy, to be honest, considering the established technological base within mankind. Joking about him would be just mean...online.
#cyberpunkcoltoure
PS:
To calculate the energy required for a laser communications link to successfully transmit a confirmation signal back to Earth, we must look at the physics of beam divergence.
Because a laser beam spreads out as it travels across 4.3 light-years, a signal that leaves the star system with high intensity will be incredibly faint by the time it reaches a collector telescope on Earth.
The energy required depends directly on how long the transmission lasts. The standard way to calculate this in aerospace engineering is to determine the required received signal energy (usually about 20 to 100 photons per bit of data) and scale it up based on the distance.
Assuming a highly optimistic, state-of-the-art 10-meter receiving telescope on Earth and a near-infrared laser (1064 nm) fired from a compact 1-meter transmitter aperture on the satellite, here is the energy breakdown to send a basic, low-data confirmation package (e.g., a few kilobytes of data):
Because a laser beam spreads out as it travels across 4.3 light-years, a signal that leaves the star system with high intensity will be incredibly faint by the time it reaches a collector telescope on Earth.
The energy required depends directly on how long the transmission lasts. The standard way to calculate this in aerospace engineering is to determine the required received signal energy (usually about 20 to 100 photons per bit of data) and scale it up based on the distance.
Assuming a highly optimistic, state-of-the-art 10-meter receiving telescope on Earth and a near-infrared laser (1064 nm) fired from a compact 1-meter transmitter aperture on the satellite, here is the energy breakdown to send a basic, low-data confirmation package (e.g., a few kilobytes of data):
## Energy Required for Confirmation Signal
| Location | Distance to Earth | Total Pulse/Transmission Energy | Equivalent Everyday Energy |
|---|---|---|---|
| Outskirts (Oort Cloud equivalent, ~1.5 light-years)
| Location | Distance to Earth | Total Pulse/Transmission Energy | Equivalent Everyday Energy |
|---|---|---|---|
| Outskirts (Oort Cloud equivalent, ~1.5 light-years)
| ~14.2 trillion km | ~5.5 × 10⁶ Joules (5.5 MJ) | A few minutes of running a household microwave. |
| Center of System (Proxima / Alpha Centauri A/B, ~4.3 light-years)
| Center of System (Proxima / Alpha Centauri A/B, ~4.3 light-years)
| ~41.1 trillion km | ~4.6 × 10⁷ Joules (46 MJ) | The energy stored in about 1.3 liters of standard gasoline. |
------------------------------
## The Power vs. Energy Catch
While the total energy (Joules) listed above sounds surprisingly small and manageable (easily stored in a small lithium-ion battery on a satellite), the physics of interstellar space creates a severe Power (Watts) problem.
* The Problem of Noise: The Alpha Centauri system features three massive, blindingly bright stars. If the satellite transmits a weak, slow signal, Earth's telescopes will completely lose the laser's photons in the overwhelming glare of the background starlight.
* The Solution: The satellite must compress that energy into an incredibly fast, ultra-intense flash. To overcome the background noise, a common strategy is to fire the signal as a 100-watt to 1-kilowatt laser beam sustained for a few hours.
## Why the Outskirts require 8x less energy
As a laser beam travels, its diameter expands linearly with distance, meaning the area of the beam's footprint expands quadratically ($Distance^2$). Because the "Center" of the system is roughly 2.8 times farther away than the "Outskirts," the laser beam expands to cover an area nearly 8 times larger by the time it hits Earth. Consequently, the satellite at the center must fire a beam that is 8 times more energetic to ensure the exact same number of photons land inside Earth's receiving telescope.
------------------------------
## The Power vs. Energy Catch
While the total energy (Joules) listed above sounds surprisingly small and manageable (easily stored in a small lithium-ion battery on a satellite), the physics of interstellar space creates a severe Power (Watts) problem.
* The Problem of Noise: The Alpha Centauri system features three massive, blindingly bright stars. If the satellite transmits a weak, slow signal, Earth's telescopes will completely lose the laser's photons in the overwhelming glare of the background starlight.
* The Solution: The satellite must compress that energy into an incredibly fast, ultra-intense flash. To overcome the background noise, a common strategy is to fire the signal as a 100-watt to 1-kilowatt laser beam sustained for a few hours.
## Why the Outskirts require 8x less energy
As a laser beam travels, its diameter expands linearly with distance, meaning the area of the beam's footprint expands quadratically ($Distance^2$). Because the "Center" of the system is roughly 2.8 times farther away than the "Outskirts," the laser beam expands to cover an area nearly 8 times larger by the time it hits Earth. Consequently, the satellite at the center must fire a beam that is 8 times more energetic to ensure the exact same number of photons land inside Earth's receiving telescope.
If you are interested, I can show you how the data rate (how many bits per second we can receive) changes based on how much power the satellite has available, or what kind of Earth-based telescope arrays we would need to build to catch the signal.
And you have to hit a receiver on Earth. Do you've got one already? You have some time so. 6.799 years give or take a few months, approximately.
#sometimesIcanthelpmyself #theregoesmyhumor #googleAI
Do we all fuck for well understand that I pulled that from a free of charge online AI, namingly Google, and they as simply 15 million U.S. Dollars into their wallets?
What am I missing??? I am always broke.
