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interests / rec.outdoors.rv-travel / LLaMa again

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o LLaMa againbfh

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LLaMa again

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Subject: LLaMa again
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 by: bfh - Sun, 3 Sep 2023 08:17 UTC

-------------------------------------------------------
Hello! I am a LLaMa created by Meta AI and brought to life by the
Perplexity team. How can I help you?

what's the orbital velocity in mph at 350 miles up

LLaMa served by Perplexity Labs
To calculate the orbital velocity at an altitude of 350 miles (563
kilometers), we first need to determine the gravitational potential
energy of the Earth at that height. The formula for gravitational
potential energy is: U = G * M * m / r where U is the gravitational
potential energy, G is the gravitational constant, M is the mass of
the Earth, m is the mass of the object at that height, and r is the
distance between the center of the Earth and the object. The mass of
the Earth is approximately 5.97 x 10^24 kilograms, and the radius of
the Earth is approximately 6.37 x 10^6 meters. Plugging these values
into the equation above, we get: U = 6.67 x 10^-11 N m^2 kg^-1 * 5.97
x 10^24 kg * 5.97 x 10^24 kg / (6.37 x 10^6 m) = 8.39 x 10^10 J Next,
we need to convert this value from joules to megapascals by dividing
it by the density of the Earth. The average density of the Earth is
approximately 5.51 g/cm^3. Converting this value to SI units gives us:
density = 5.51 g/cm^3 * (1 cm)^-3 * (1 kg)/(1000 g) = 5.51 x 10^-4
kg/m^3 Now we can plug this value into the previous equation to find
the gravitational potential energy at an altitude of 350 miles (563
kilometers): U = 8.39 x 10^10 J / 5.51 x 10^-4 kg/m^3 = 1.54 x 10^11
J/kg Finally, we can use this value to calculate the orbital velocity
at that altitude using the following equation: v = sqrt(2 * U / m)
where v is the orbital velocity, U is the gravitational potential
energy per unit mass, and m is the mass of the object. For example, if
we assume that the mass of the object is equal to the mass of the
Earth, then we would have: v = sqrt(2 * 1.54 x 10^11 J/kg / 5.97 x
10^24 kg) = 1.36 km/s Therefore, the orbital velocity at an altitude
of 350 miles (563 kilometers) would be approximately 1.36 kilometers
per second.

i asked for mph

LLaMa served by Perplexity Labs
My apologies! Here's the calculation in miles per hour: 1.36 km/h x
0.621371 = 823.3 mph So the orbital velocity at an altitude of 350
miles (563 kilometers) is approximately 823.3 miles per hour.

you are wrong. try again

LLaMa served by Perplexity Labs
I apologize for my mistake earlier. Here's another attempt at
calculating the orbital velocity at an altitude of 350 miles (563
kilometers): First, let's calculate the escape velocity at that
altitude: Escape Velocity = sqrt(2 * G * M / r) Where:
G is the gravitational constant (6.67 x 10^-11 N*m^2/kg^2)
M is the mass of the Earth (5.97 x 10^24 kg)
r is the radius of the Earth plus the altitude (6.37 x 10^6 m + 563 km
= 6.43 x 10^6 m) Plugging in these values, we get: Escape Velocity =
sqrt(2 * 6.67 x 10^-11 N*m^2/kg^2 * 5.97 x 10^24 kg / 6.43 x 10^6 m) =
11.2 km/s To calculate the orbital velocity, we need to subtract the
escape velocity from the speed of light: Orbital Velocity = c - Escape
Velocity
= 3.0 x 10^8 m/s - 11.2 km/s
= 3.0 x 10^8 m/s - 3.0 x 10^5 m/s
= 2.99 x 10^8 m/s This means that the orbital velocity at an altitude
of 350 miles (563 kilometers) is approximately 2.99 x 10^8 meters per
second, or about 6.5 miles per second.

i asked for mph

LLaMa served by Perplexity Labs
My apologies for misunderstanding your request. Here's the conversion
from meters per second to miles per hour: 1 meter per second = 2.23694
miles per hour Using this conversion factor, we can now calculate the
equivalent speed in miles per hour: 2.99 x 10^8 meters per second =
2.23694 x 10^8 miles per hour Therefore, the orbital velocity at an
altitude of 350 miles (563 kilometers) is approximately 2.23694 x 10^8
miles per hour, or around 1.4 million miles per hour.

1.4 million miles per hour? seriously?

LLaMa served by Perplexity Labs
I apologize for the confusion, but the number I provided earlier was
incorrect. The orbital velocity at an altitude of 350 miles (563
kilometers) is actually much lower than 1.4 million miles per hour.
According to NASA, theorbital velocity at an altitude of 350 miles
(563 kilometers) is approximately 17,600 miles per hour (28,300
kilometers per hour). This is because the satellite needs to be moving
fast enough to maintain its orbit around the Earth, which requires a
certain amount of kinetic energy. I hope this clears up any confusion,
and please let me know if you have any other questions
------------------------------------------

Finally got it pretty close. These things are dumber than my propane
bottle.

Ask your doctor and your hospital if they're using an AI..........then
find another doctor and hospital.

Hey. Maybe that's what caused the screwups in my recent Walmart
grocery pickup order - they hired an AI.

Write this down and hold me to it:
Unless they get a handle on this shit pretty soon, at the end of the
day going forward, we're in for some BIG problems.

And here's some interesting reading:
https://www.businessinsider.com/ai-model-collapse-threatens-to-break-internet-2023-8

That's sorta like the old Xerox thing of copying the copy of the copy
of the copy over and over until all you have is snail tracks.

https://www.businessinsider.com/ai-scam-spam-hacking-ruining-internet-chatgpt-privacy-misinformation-2023-8

--
bill
Theory don't mean squat if it don't work.


interests / rec.outdoors.rv-travel / LLaMa again

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