Mass of the Earth Traveling at 500,000 mph?

[Uberhulk]

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]

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]

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.

The mass of the planet would only change if it was traveling beyond FTL?

 

[russ_watters]

The mass of the planet would only change if it was traveling beyond FTL?

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).

Also, there's no traveling FTL, much less "beyond FTL", whatever that means...though since this is the science fiction section, I guess it can mean whatever you want.

 

[Ibix]

As Russ notes, nobody (edit: almost nobody, anyway) uses relativistic mass anymore because it leads to enormous confusion - it isn't a mass in all the senses of the word. So the first question would be why do you want to know?

Secondly, relativistic mass (if that turns out to be what you want) is a factor of $1/\sqrt{1-v^2/c^2}$ larger than rest mass. For a speed of half a million miles an hour, about 0.0007c, that is about 1.0000003, about three parts in ten million larger. That's more or less completely negligible.

 

[Uberhulk]

Ok. Let's look at this another way. What force (in Newtons) would be required to stop a planet with Earth's mass traveling at 500,000 mph?

 

[Ibix]

Depends how long you want to wait. And whether you want it in one piece afterwards or not.

Stop it in 1s? Very very high. Stop it in a couple of million years? High. If you're happy to wait long enough, though, you could do it with a bottle rocket, if you could make it sustain thrust long enough.

 

[Uberhulk]

Stopping it by catching it, like this:

 

[Ibix]

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.

 

[Bandersnatch]

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.

 

[Uberhulk]

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.

Yes, it's Hyperion, he survived the collapse of two universes so it's in keeping with his strength and durability, I guess.

 

[Uberhulk]

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.

Thank you. Yes, comic book physics! The planet was traveling at 500,000 mph, as shown in an earlier panel. I'm wary of posting images on this forum.

 

[PeroK]

Yes, it's Hyperion, he survived the collapse of two universes so it's in keeping with his strength and durability, I guess.

A true superhero is not going to let himself be constrained by the laws of physics!

 

[Vanadium 50]

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?

 

[Ibix]

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?

The planet rushes onward - but space ITSELF accelerates ahead of it, driven before the MIGHTY forces wielded by this superhero!

(I dunno, but physics has nothing to do with it, so might as well have some fun with it...)

 

[256bits]

'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.

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.

 

[Uberhulk]

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? Superman benched the weight of Earth, Thor lifted a score of planets and Hulk overcame the weight of a star, when I have the calculation showing the power needed to lift one planet, I can then calculate the other two figures.

Yes, I know the physics doesn't work and the planets would be crushed under the weight!


[mentors’ note - copyrighted images have been removed. If this is a mistake and the images are in fact legally available, PM one of the mentors and we can restore them]

How Big is the Sun? | Size of the Sun

The total volume of the sun is 1.4 x 1027 cubic meters. About 1.3 million Earths could fit inside the sun. The mass of the sun is 1.989 x 1030 kilograms, about 333,000 times the mass of the Earth.

 

[Vanadium 50]

Yes, I know the physics doesn't work and the planets would be crushed under the weight

Then what is the point of this thread?

 

[PeroK]

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?

The concept of "lifting" something depends on the presence of a large mass (like a planet) with a gravitational field. It then takes energy to lift something, which means move it to a higher potential in the field.

If you are on the Earth, there is no concept of "lifting" the Earth against its own gravity.

If, say, the Moon was sitting on the surface of the Earth, then the force to lift the Moon would depend not only on the Moon's mass, but the mass of the Earth, against whose gravity you are doing the lifting.

If you want to lift the Earth, where are going to put your feet?

 

[Uberhulk]

@Vanadium 50 This is the Science Fiction and Fantasy Media forum. You may as well ask what's the point of any thread in this forum.

 

[PeroK]

This is the Science Fiction and Fantasy Media forum.

To lift the mass of the Earth above your head, all you have to do is a handstand!

 

[Uberhulk]

If you want to lift the Earth, where are going to put your feet?

In these feats, they are standing on Earth and taking magic and cosmic power equivalent to the weight of 20 planets and equivalent to the weight of star.

 

[Vanadium 50]

. You may as well ask what's the point of any thread in this forum.

I'm certainly asking what the point is for this thread, where you admit the physics is just made up.

 

[Uberhulk]

I'm certainly asking what the point is for this thread, where you admit the physics is just made up.

Of course I admit the physics is made up! This is the sci-fi and fantasy forum! The physics in most of the topics here are made up - there's a question about time travel, about Star Wars, compression in Pokeballs etc, etc so why are you asking? Science Fiction and Fantasy take elements from physics but they do not obey the laws of physics, everyone knows this. What do you expect in these forums??

Comic books fans have been using physics for decades to determine energy levels or levels of force, for the purpose of debate, to determine the power levels of characters, to determine is A is stronger than B?

 

[DaveC426913]

I think, a the very least, the question must be well-formed. In this case:

how much is required to lift the mass of one planet, if that planet has the same mass as Earth?

"lift" is ambiguous.

If you're standing on the Earth, how can you lift it?

Best scenario I can make out of this is if you have two Earth-sized planets in contact, and our hero is standing between them. Now, that means you're dealing with the attraction of two Earth masses, not just one.

Also, you'll have to hand-wave away the effects of the Roche Limit - which will result in the two planets disintegrating to form a single mass of rubble.

 

[DaveC426913]

Hey thanks @Uberhulk ! You're my 2000th positive reaction! I just got the award!

 

[Uberhulk]

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?

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

I termed the question that way for a reason. In the two feats I showed, neither Hulk or Thor are lifting a physical object, they're been hit with the mass equivalent to 20 planets and a star. This is how Superman benched the weight of the Earth, using a giant machine that simulated the weight:

New 52 Superman Benches Weight Of Earth for Five Days. Superman 13 (2012)

http://www.ecology.com/2008/09/08/how-much-does-earth-weigh/

Earth's weight is 5.972 sextillion (1,000 trillion) tons or 5,972,000,000,000,000,000,000. A thousand to the power of 21 (10 x 21).

[mentors’ note - copyrighted images have been removed.]

 

[DaveC426913]

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

I know, but "lift" is still ambiguous in that context.

 

[jbriggs444]

Now, that means you're dealing with the attraction of two Earth masses, not just one.

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.

 

[DaveC426913]

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.

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?

 

[Janus]

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?

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.

 

[jbriggs444]

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.

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.

 

[Janus]

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.

I was considering a "Earth equivalent" mass being supported at the surface of the Earth, not two Earth-sized objects.