# Does velocity affect Gravity.

## Main Question or Discussion Point

Based off the special theory of relativity the faster an object moves then the more massive it becomes (as long as it is still under the speed of light). So based off that and the universal law of gravitation, if an object becomes more massive then does the Earth apply a stronger gravitational force to it (and it to the Earth) or does the Earth and it always apply the rest mass amount of gravitational force?

For example, if a bullet where hypothetically traveling at 99% of the speed of light just above the Earth's surface then its mass would be 7 times greater than its rest mass. Besides drag would the vertical component of the force (force of gravity) on the bullet be 7 times more than it would be if the bullet was just dropped from rest just above the Earth's surface?

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Based off the special theory of relativity the faster an object moves then the more massive it becomes (as long as it is still under the speed of light).
This is not physical and does have no experimental backup to rely on it!

So based off that and the universal law of gravitation, if an object becomes more massive then does the Earth apply a stronger gravitational force to it (and it to the Earth) or does the Earth and it always apply the rest mass amount of gravitational force?
This is the third law of Newton which says the more you are massive, the more you force the ground, thus the stronger the reactive force is! If you have a mass $$m$$ and $$M$$ is the Earth mass, and the distance between you and earth is $$r-r_0$$, the Newtonian law of gravitation says that the gravitational force is $$\textbf{F}=-Mm\hat{r}/(r-r_0)^3.$$ You can see if $$m$$ gets larger, the magnitude of $$\textbf{F}$$ increases!

For example, if a bullet where hypothetically traveling at 99% of the speed of light just above the Earth's surface then its mass would be 7 times greater than its rest mass. Besides drag would the vertical component of the force (force of gravity) on the bullet be 7 times more than it would be if the bullet was just dropped from rest just above the Earth's surface?
You don't get massive in reality when traveling at a speed close to c. It may be true to say if a bullet is hypothetically traveling at 99% of the speed of light and suddenly hits the surface of the Earth, then the damage it causes must be much more than when it hits the Earth's surface with a lower speed. But this does not have anything to do with gravity at all! The gravitational force is only proportional to the mass of object and the inverse squared distance between object and the gravitating body.

AB

In Einstein's free falling elevator thought experiment he demonstrates the principle of equivalence by showing that to an observer inside the elevator, everything inside the elevator appears to behave as if the elevator was was far out in space with no acceleration acting on it at all. From the point of view of an observer inside the free falling elevator, a bullet fired horizontally across the elevator travels in a straight line and hits the same point on the wall as when the elevator is not accelerating in flat space. This demonstrates that the bullet drops at the same rate as the elevator. The same is true if he shines a laser beam across the elevator. The light falls at the same rate as a ball released from his hand. The light example shows that even an particle with zero rest mass drops at the same rate as everything else and in fact in General Relativity gravity is not thought of in terms of gravity acting as a force on mass.

The elevator example assumes that you have a very small elevator falling towards a very large planet to eliminate tidal effects that will change things slightly.

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Yes, the grav field of an object moving near the speed of light is greater than when it is at rest. The moving object has a huge amount of kinetic energy. And energy produces gravity, just like mass does. E = MC^2

bcrowell
Staff Emeritus
Gold Member
I don't think kev's answer in #3 is quite right. The equivalence principle requires that everything have the same downward acceleration. However, that doesn't mean that everything has to feel the same force. An object moving horizontally with $\gamma=7$ has 7 times its normal inertia, but it also has 7 times the normal amount of gravitational force acting on it, so it has the same acceleration it normally would have.

Although it's true that GR doesn't really describe gravity as a force, the correspondence principle guarantees that we should be able to interpret it that way in the appropriate Newtonian limit. The gravitational attraction between the earth and an apple is certainly affected by the energy content of each body, as given by E=mc2.

Once you go to sufficiently high order corrections beyond the Newtonian limit, you will get more complicated stuff happening. The concept of force will not work as well. Also, you can get nonlinear effects, so that bodies don't necessarily travel along geodesics. (It's only low-mass particles that travel along geodesics in GR. Once you give an object enough mass-energy, all kinds of effects can happen, e.g., it can emit gravitational waves.)

bcrowell
Staff Emeritus
Gold Member
Yes, the grav field of an object moving near the speed of light is greater than when it is at rest. The moving object has a huge amount of kinetic energy. And energy produces gravity, just like mass does. E = MC^2
Yes, but as others here have pointed out to me before, in general you can't just apply E=mc2 and get the field of a moving body accurately based on the mass equivalent of its kinetic energy, m+K/c2. In a frame where the body is seen as moving, its stress-energy tensor doesn't just have a $T_{tt}$, it has other elements as well, so the field isn't a Schwarzschild metric. I suspect that the metric would be approximately a Schwarzschild metric for m+K/c2 in some average sense or in some appropriate limit.

Yes, but as others here have pointed out to me before, in general you can't just apply E=mc2 and get the field of a moving body accurately based on the mass equivalent of its kinetic energy, m+K/c2. In a frame where the body is seen as moving, its stress-energy tensor doesn't just have a $T_{tt}$, it has other elements as well, so the field isn't a Schwarzschild metric. I suspect that the metric would be approximately a Schwarzschild metric for m+K/c2 in some average sense or in some appropriate limit.
So you are saying that it is not entirely correct to set up a differential equation

d(m(v)*v)/dt=-G*M(v)*m(v)/R^2+Drag forces

to describe a very fast projectile.

I'm doing a project where a projectile is going very fast. The differential equation I set up is

d(mv)/dt = drag force equation - G*M*m0/(R+h)^2 where m=m0/sqrt(1-v^2/c^2) is the projectiles mass, M is the mass of the Earth, R is the radius of the Earth, and h is the distance above the Earth's surface.

But I was thinking to use m instead of m0 in the second term of the right side (gravity part).

I don't think kev's answer in #3 is quite right. The equivalence principle requires that everything have the same downward acceleration. However, that doesn't mean that everything has to feel the same force. An object moving horizontally with $\gamma=7$ has 7 times its normal inertia, but it also has 7 times the normal amount of gravitational force acting on it, so it has the same acceleration it normally would have.
Hi Ben,
I admit I was focusing more on the acceleration issue and whether the bullet with high horizontal velocity drops at the same rate as a bullet with zero horizontal (i.e. they hit the ground at the same time) rather than on the force issues that the OP is probably more interested in.

I see we are in agreement that the downwards acceleration is the same as per the equivalence principle.
Although it's true that GR doesn't really describe gravity as a force, the correspondence principle guarantees that we should be able to interpret it that way in the appropriate Newtonian limit. The gravitational attraction between the earth and an apple is certainly affected by the energy content of each body, as given by E=mc2.
The force issue is a little more tricky and there is more to it than simply E=mc2 and we have to be careful that our conclusions do not imply to newcomers that an object with high relative velocity is in danger of collapsing into a black hole, which we know is not the case.

I came up with this equation for the gravitational force acting on a test object which takes the horizontal velocities of the gravitationally massive object and the test particle into account.
$$F' = \frac{GMm}{R^2} \frac{(1-Vv/c^2)^2}{(1-V^2/c^2)\sqrt{1-v^2/c^2}}$$

where V is the velocity of the massive body (M) wrt to the observer and v is the velocity of the test particle (m) with respect to the observer.

It is assumed here that M is much greater than m so that we do have to be concerned with acceleration of M towards m.
My equation is intended for use in the weak gravitational limit (ie Earth scale gravity rather than black hole scale gravity) and agrees with your assertion that when an object has horizontal velocity v that it feels an increased downward force by a factor of 1/sqrt(1-v^2/c^2) and I would tend to agree that the increased transverse inertia cancels out the increased force so that downward acceleration ends up the same.

When the test particle and the gravitational body are co-moving (V=v) the gravitational force on the particle is reduced by a factor of gamma so there is no danger of gravitational collapse to a black hole. It is known from Special Relativity that transverse static forces are reduced by a factor of gamma so the internal forces supporting the the gravitational body remain in equilibrium with the gravitational forces. However it should be noted that in this case the test particle and the gravitational body both have increased energy according to E=mc2 but there is no corresponding increase in gravitational force.

In the case of V=v the downward acceleration of the test particle is reduced by a factor of gamma squared (ie 49 times slower than when V=v=0). This is agreement with SR because an observer comoving with the two bodies will measure the downward acceleration as normal because of the slower rate of their clock.

It can also be worked out that if the bullet fired from the gun and the dropped bullet hit the ground simultaneously according to an observer at rest with the ground, then to an observer co-moving with the bullet the fired bullet drops faster than the dropped bullet and hits the ground before the dropped bullet due to the lack of absolute simultaneity in SR.

Now here is an interesting thought experiment. Imagine a train on a long straight horizontal monorail that is suspended from springs that directly measure the weight of the train and the rail. When the train is accelerated to the same velocity as the bullet in the OP will the track scales directly measure the train to be 7 time heavier?

I'm doing a project where a projectile is going very fast. The differential equation I set up is

d(mv)/dt = drag force equation - G*M*m0/(R+h)^2 where m=m0/sqrt(1-v^2/c^2) is the projectiles mass, M is the mass of the Earth, R is the radius of the Earth, and h is the distance above the Earth's surface.

But I was thinking to use m instead of m0 in the second term of the right side (gravity part).
You have to be careful here with small spherical gravitational bodies like the Earth and serious relative velocities such as 0.99c because you have to take into account issues with the projectile effectively going into orbit and not falling to the ground at all. The projectiles still fall relative to an imaginary straight line tangential to a point on the ground level with the launch point, but you should bear that factor in mind.

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When the test particle and the gravitational body are co-moving (V=v) the gravitational force on the particle is reduced by a factor of gamma so there is no danger of gravitational collapse to a black hole. It is known from Special Relativity that transverse static forces are reduced by a factor of gamma so the internal forces supporting the the gravitational body remain in equilibrium with the gravitational forces. However it should be noted that in this case the test particle and the gravitational body both have increased energy according to E=mc2 but there is no corresponding increase in gravitational force......
.......Now here is an interesting thought experiment. Imagine a train on a long straight horizontal monorail that is suspended from springs that directly measure the weight of the train and the rail. When the train is accelerated to the same velocity as the bullet in the OP will the track scales directly measure the train to be 7 time heavier?
The original question & the thought-experiment suggested are both tricky to answer with Special Relativity. If I'm not mistaken,you've assumed the equality of inertial & gravitational mass implicitly (which is unaccountable with Special Relativity ). Although the static force experienced by a test particle changes , the field parameter (electric charge, for e.g.) may not change.
If one should define gravitational field as usual ,i.e.,
force experienced / rest mass = 'field strength' , the field strength would depend on the velocity. We don't get over this by changing the rest mass by 'relativistic mass' (unless the field is radial & the motion is entirely transverse or radial)or even with the corresponding 4-vectors.I don't know how to define the field. As for the original problem, we can put the longitudinal or transverse mass in the Newton's law of gravity ; provided the motion is according.
By General relativity, of course,a test particle moves along a geodesic regardless of its little mass . As the speed grows, however, the motion becomes complicated ( it 'changes the field', if you'd like a classical mnemonic).

Hi Ben,

You have to be careful here with small spherical gravitational bodies like the Earth and serious relative velocities such as 0.99c because you have to take into account issues with the projectile effectively going into orbit and not falling to the ground at all. The projectiles still fall relative to an imaginary straight line tangential to a point on the ground level with the launch point, but you should bear that factor in mind.
I really want to see how much force or which what velocity would it take to hit Superman into space. I understand the drag forces of the equation but I'm not sure if the gravitational field is GMm/R^2 or GMm/R^2*1/sqrt(1-v^2/c^2) where v is the velocity of the object with rest mass m.

I really want to see how much force or which what velocity would it take to hit Superman into space. I understand the drag forces of the equation but I'm not sure if the gravitational field is GMm/R^2 or GMm/R^2*1/sqrt(1-v^2/c^2) where v is the velocity of the object with rest mass m.
A surface skimming Earth orbit (ignoring the atmosphere) has a velocity of about 7.8 km/s and for circular orbits with a larger radius the orbital velocity is even less because orbital velocity = sqrt(GM/r). The escape velocity from the surface of the Earth is about 11.2 km/s and anything moving with that velocity or greater escapes to infinity. The speed of light is about 300,000 km/s and 11.2 km/s is a tiny fraction of that so it is not possible to have orbital speeds where relativistic effects are significant.

If you want to accelerate something to relativistic velocities using something like a particle accelerater the force required is gamma^3 times the Newtonian force where gamma = 1/sqrt(1-v^2/c^2).

A surface skimming Earth orbit (ignoring the atmosphere) has a velocity of about 7.8 km/s and for circular orbits with a larger radius the orbital velocity is even less because orbital velocity = sqrt(GM/r). The escape velocity from the surface of the Earth is about 11.2 km/s and anything moving with that velocity or greater escapes to infinity. The speed of light is about 300,000 km/s and 11.2 km/s is a tiny fraction of that so it is not possible to have orbital speeds where relativistic effects are significant.

If you want to accelerate something to relativistic velocities using something like a particle accelerater the force required is gamma^3 times the Newtonian force where gamma = 1/sqrt(1-v^2/c^2).
relativistic velocities do come into play when not ignoring the atmosphere. Actually to project something straight upward in orbit takes many times more velocity and can affect the the outcome of the force required to project something into orbit significantly. For example, I can calculated accurately that it takes millions of tons of force (using only 2m distance of acceleration) to project somethings into orbit. But adding in drag forces from an atmosphere and we get astronomically more force needed (much higher initial velocities). What is the Newtonian force? Is gamma^3 times it the gravitational force?