Two clocks synchronized by a light pulse in the "train frame":
The same clocks according to the ground frame,
Trailing clock starts first.
P.S. Time dilation and Length contraction not accounted for in these animations.)
The GPE equation above give the value relative to a point an infinite distance from the source of gravity ( where GPE is considered zero). What you want is the difference in GPE between two points.
The orbit is nearly 4.5 times further from the center of the planet than the surface. What happens to the value of the acceleration due to gravity as you move away from the planet?
So, jut to clarify what the last two two posters have said. If in the ship frame, the light pulses reaches a clock when it reads 12:00:01, then according to the Earth frame, this is exactly what that clock reads when the light pulse reaches it. In the ship frame, both clocks read the same...
So, they should arrive, when the clocks at the ends read the same.
What time should each clock read when its light pulse reaches it in the Earth frame?
Your answers are off by a few magnitudes. The Earth-Moon distance is 3.84e8 m, not 3.84e6m. Thus your answer is 10,000 times too large.
First, I assume you meant Fg in this last equation. As written, it just simplifies to v=v. Secondly, from the answers below, I is apparent that you took r...
You didn't get this right. After 21.1 ns, the light will have indeed traveled 6.34 m. However, in 21.1 ns, the leading mirror will have moved 3.8 m, and will now be 7.8 m from where the light left the trailing mirror. In other words, it will still be 1.46 m ahead of the light.
This is...
You forgot to properly account for the fact that, in the galaxy frame, the mirrors are moving in the x direction.
So, for example, Light leaving the trailing mirror, after 13ns will have traveled a distance of 4 m in the x direction ( the distance between the mirrors as measured in the galaxy...
gneill is pointing you in the right direction. While the answer you get for the speed of the rocket at 970 km seems about right, the velocity vector won't be pointing the right direction when you get to that altitude, so it isn't going just be a matter of increasing speed to match the orbital...
It looks like you are trying to use the equation for escape velocity instead of that for orbital velocity. And even if you use orbital velocity, this will not give a good estimate.
As far as working with angular momentum is concerned: Yes, you are changing the orbit, so while the angular...
It's probably easier to see what happening with an an animation.
First from the frame of the ship. The small dot represents the laser path. it starts from the ship when the the laser reaches the edge of the cloud and the point of the cloud the laser is at when it returns is marked off by the...
The issue I see here is that for the laser fired ahead of the ship you have to take the relativity of simultaneity into account. The reason you get different answers for the cloud frame vs the ship frame is that the two frame measure simultaneity differently. Let's call the point of the...
The thing to remember is that which observer is the "Stationary" observer depends on which frame you are working from. In this scenario, If you are measuring from the Ground frame, then the ground observer is stationary and the man is moving. But in the Man's frame, he is stationary and the...
According to the question, 450 km is the distance between Montreal and Toronto as measured by the man flying between them.
Yes, you are correct, the proper distance between Toronto and Montreal as measured in the ground frame would be longer than that measured by the man. (However, I would...
At the moment the Sun loses mass, the Earth is a certain distance from the Sun (r) and has a given velocity(v). The Sun losing mass will not change these properties at that moment, just the future trajectory of the Earth from that moment on.
If they were perfectly rigid, then you could not say that they spent any measurable time (fraction of a second) in contact. All changes in velocities would be instantaneous. The ball would instantaneously change direction and the bat would instantaneously lose a bit of velocity on contact...
Starting with an initial velocity can be handled by using energy conservation to simplify to a zero initial velocity. In the upward moving cup scenario, you solve for how high the cup will climb before reaching the apex, and then solve for the apex to starting position fall time.
If you...
You can attack this in two parts:
First, by energy conservation, work out how high your cup travels above 1000km before it starts to fall back.
Then you can solve for the time it takes to fall from a point at rest from that altitude to 1000 km (there is an implicit formula for this), keeping in...
One method is the loop-equation method. Draw out the possible current loops for the circuit, write an equation for each loop, and then solve the simultaneous equations to find each loop current. The sum of these give you the total current for any given source voltage, and from that you can...
Point 1: If Ki+Ui is the initial total energy of the object while at rest with respect to the Earth and Kf+Uf is the final energy of the object in orbit, Then why would you add energy to the total final energy to get the total initial energy?
Point 2: Just relating initial energy, added...
Question b is quite straight forward if you think about it. It basically breaks down to this:
What happens to the time dilation formula if you substitute 2v for c? (note that you don't actually have to know the value of v to do this.)
Since nothing is stated in the problem regarding the altitude of the orbiting clock (which also determines v), it seems he simply chose what he felt was a convenient value. The problem with the value he chose was that it put the orbiting clock at a higher altitude than the Earth surface clock...
Just splitting the spacecraft into two unequal masses will not cause the two parts to follow different trajectories. When you are dealing with a relative mass ratio like you get between a spacecraft and a planet, the mass of the spaceship (or its separate parts) has no effect on the orbital...
A "solar day" is the time it take for the Sun to go from noon to noon as seen from a given spot on the Earth. It is the "Earth day" upon which we base our clocks.
What was being pointed out was that this would be equal to length of time that an observer on the Sun would see between two instances...
You calculated the acceleration due to gravity, not the acceleration of the ball due to the force of the spring. One acts downward and the other upward.
Not quite what I was looking for. There is a fairly direct way to get the answer to the problem, but it assumes some knowledge of the nature of orbits. I was trying to gauge just how much you knew about orbits in order to know in what direction to lead you in.