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Uberhulk
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What would be the mass be of a planet traveling at 500,000 mph - if we assume the planet has the same mass as the Earth i.e. 5.9 sextillion tonnes?
I assume you're asking about relativistic effects? That speed is less than 1/1000th the speed of light, so relativist effects can be ignored.Uberhulk said:What would be the mass be of a planet traveling at 500,000 mph - if we assume the planet has the same mass as the Earth i.e. 5.9 sextillion tonnes?
russ_watters said:I assume you're asking about relativistic effects? That speed is less than 1/1000th the speed of light, so relativist effects can be ignored.
First a caveat that scientists don't really use the "mass increase" idea anymore. But no, the effects are always there, they are just generally considered negligible until you are moving a substantial fraction of the speed of light (like half).Uberhulk said:The mass of the planet would only change if it was traveling beyond FTL?
Bandersnatch said:So, how long is that supposed to take then? Like ten seconds, maybe?
You then have ##F=m\Delta V/\Delta t##
with ΔV = 500,000 mph = 805,000 km/h = 224,000 m/s (assuming by panel 2 the planet has been stopped entirely)
Δt = 10s
m = 6*10E24 kg (like Earth)
which gives F=134*10E27N, so let's round it to F=10E29N for simplicity
Now, it looks like he's trying to stop it with his hands and chest. On average, hands have total surface area of about 0.1 square metres, chest is maybe additional 0.4. Let's say it's twice that because he's a huge buffed-up superhero, so 1 m^2 total. This is where all of that force is concentrated, netting the pressure P=10E29Pa.
This is something like 20 orders of magnitude above the strength of any material in existence (we're assuming here the superdude (Hyperion, is it?) is made of indestructible magical material, but nothing else is).
So if trying to perform such feat, the result would be the hero passing through the planet like a bullet through a balloon.
If he wanted to stop the planet without burrowing into and through it, he'd have to reduce the pressure exerted to something on the order of maybe 10 Mpa, so about 10,000,000 N.
This in turn means that the time between panels 1 and 2, between the 'Hurk!' and 'I... I have it', would have to be extremely compressed, since at that force it'd take in excess of 3 trillion years - longer that the age of the universe.
Ibix said:So he's stopping a mass of ##6\times 10^{24}kg## moving at something like ##2\times 10^5m/s## in something like 20s (estimating from the dialog). That's an acceleration of about ##10^4m/s^2##. Newton's second law tells us that force is mass times acceleration, so about ##6\times 10^{28}N## required.
Note that he's violating conservation of momentum doing what he's doing, so this number isn't really relevant because physics clearly has nothing to do with this story. Also, a thousand gravities would smash the planet. So there's yet more magic going on here.
A true superhero is not going to let himself be constrained by the laws of physics!Uberhulk said:Yes, it's Hyperion, he survived the collapse of two universes so it's in keeping with his strength and durability, I guess.
The planet rushes onward - but space ITSELF accelerates ahead of it, driven before the MIGHTY forces wielded by this superhero!Vanadium 50 said:Taking a planet's velocity from 500,000 mph to zero is the same as taking it from zero to 500,000, just in a different coordinate system. If your hero stands on his hands and pushes, what exactly is supposed to happen?
since it's Hyperion. so perhaps he likes to put things in a HYPERbolic orbit ( hence his name! ), so a little bit of a nudge to the left or right would avoid a catastrophic collision, and the compression of time would necessarily be reduced somewhat.Bandersnatch said:'I... I have it', would have to be extremely compressed, since at that force it'd take in excess of 3 trillion years - longer that the age of the universe.
Uberhulk said:Yes, I know the physics doesn't work and the planets would be crushed under the weight
Uberhulk said:I have a related question. Taking speed out of the equation, if we assume a planet is stationary, in terms of raw power, how much is required to lift the mass of one planet, if that planet has the same mass as Earth?
Uberhulk said:This is the Science Fiction and Fantasy Media forum.
PeroK said:If you want to lift the Earth, where are going to put your feet?
Uberhulk said:. You may as well ask what's the point of any thread in this forum.
Vanadium 50 said:I'm certainly asking what the point is for this thread, where you admit the physics is just made up.
"lift" is ambiguous.how much is required to lift the mass of one planet, if that planet has the same mass as Earth?
DaveC426913 said:I think, a the very least, the question must be well-formed. In this case:"lift" is ambiguous.
If you're standing on the Earth, how can you lift it?
I know, but "lift" is still ambiguous in that context.Uberhulk said:Not lift the Earth, lift the mass equivalent to the weight of the Earth, as I wrote in my question:
how much is required to lift the mass of one planet
No. Just one.DaveC426913 said:Now, that means you're dealing with the attraction of two Earth masses, not just one.
Doesn't the mass of two Earths create twice as strong an attraction as one Earth? Is that not why - when we measure the acceleration of things on Earth - we must drop a test object of negligible mass, so that its own gravity does not skew the measurement?jbriggs444 said:No. Just one.
For the same reason that the tension in a tug rope is equal to the tension supplied at each end. Actually, with the [counter-factual] assumption of rigid and spherical Earths, the attraction is equal to the Earth weight of 1/4 of an Earth due to the inverse square law.
Force attracting two Earth masses with their centers 1 Earth radius apart ~ 6.673e-11 (6e24)^2/6378000^2 = ~5.9e25 N. Force needed to accelerate one Earth mass at 9.81 m/sec/sec ~ 6e24* 9.81 = 5.9e25 N.DaveC426913 said:Doesn't the mass of two Earths create twice as strong an attraction as one Earth? Is that not why - when we measure the acceleration of things on Earth - we must drop a test object of negligible mass, so that its own gravity does not skew the measurement?
Two adjacent Earths with radius equal to one Earth radius will have their centers two Earth radii apart.Janus said:Force attracting two Earth masses with their centers 1 Earth radius apart ~ 6.673e-11 (6e24)^2/6378000^2 = ~5.9e25 N. Force needed to accelerate one Earth mass at 9.81 m/sec/sec ~ 6e24* 9.81 = 5.9e25 N.
I was considering a "Earth equivalent" mass being supported at the surface of the Earth, not two Earth-sized objects.jbriggs444 said:Two adjacent Earths with radius equal to one Earth radius will have their centers two Earth radii apart.
Agreed, though that the force is not doubled. Newton's third law still holds.
The mass of the Earth is calculated using the law of universal gravitation, which states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. By measuring the gravitational force between the Earth and other objects, such as the moon and satellites, scientists can calculate the mass of the Earth.
The mass of the Earth is approximately 5.97 x 10^24 kilograms. This value is constantly being refined as new measurements and data are collected.
The Earth's mass plays a crucial role in determining its orbit around the sun. The greater the mass of an object, the greater its gravitational pull. This means that the Earth's mass helps to keep it in a stable orbit around the sun, as the gravitational force between the two objects keeps the Earth from flying off into space.
The Earth's speed of 500,000 mph is its orbital velocity, which is the speed at which it orbits around the sun. This high speed is necessary for the Earth to maintain a stable orbit and remain in balance with the gravitational forces of the sun and other celestial bodies.
Yes, the Earth's mass is considered to be constant. While small changes may occur due to factors such as meteorite impacts or the addition of space debris, these changes are negligible in comparison to the Earth's overall mass. The Earth's mass has remained relatively constant throughout its history.