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actually , is is exactly the case of mass probe between two other bodies. there is no centrifugal force the gravitational time dilation calculation REQUIRES the potential above [latex]\Phi(x)=\frac{Gm}{d-x}+\frac{GM}{x}[/latex] The ratio of clock frequencies is: [latex]\frac{f®}{f(x)}=\sqrt{\frac{\Phi(x)}{\Phi®}}[/latex] The extremum occurs where [latex]\frac{d \Phi}{dx}=0[/latex] This happens for [latex]x=d(1-\sqrt{\frac{m}{M}})[/latex] This works out to be [latex]93 \sqrt{3}*10^3[/latex] miles from the center of the Earth. If you want to take into consideration the rotation of the Earth, the problem becomes a lot more complicated since the above formula for ratio of clock frequencies no longer applies, you will need to redo the derivation starting from the Schwarzschild metric in order to account for any kinematic effects. yep, my bad

Actually, it is correct. The Force acting on the test probe of mass [latex]m_p[/latex] is: [latex]\frac{Gmm_p}{(d-x)^2}-\frac{GMm_p}{x^2}[/latex]

Yes, you are correct. x would fall inside the Earth, meaning that , in fact, the potential doesn't go to 0 anywhere between the Earth and the Sun.

According to this logic, all the matrices in the exercise would have resulted to being 0. Which, according to the answers to the exercise, they clearly aren't. Sometimes , you need to be able to read BETWEEN the lines. I didn't treat anyone "like idiots", I put in a lot of effort trying to help the OP.

I have not replaced differentiation with division, I have a PhD in Physical Mathematics and I teach at a very famous university. Apparently you do not understand the notion of "convention". The same way two vectors are "divided" by taking the partial differences of their components (if the vectors are variable) one can define the vector division for the case of constant vectors by dividing them component by component. You also fail to understand the notion of "educated guess". I explained (several times) that I reversed engineered the convention from the results of the exercise.

I made it quite clear that the OP needs to go back and find where this exercise comes from. Since his vectors are constants, the only possible interpretation is that differentiation is reduced to division, this was an educated guess confirmed by the results. It also happens that if you made the effort to perform the calculations, you would have found that the resultant matrices match the answers in the linked pdf. . You weren't: I had already explained what vector/matrix differentiation is and I also explained the educated guess I was making for the OP particular case.

Yes, with this definition, you are correct, see the exact calculations in post 6

Yep, you are correct, I fixed it. Thank you for the catch, this is what happens when you cite formulas from memory, serves me right.

the distance between the Earth and the Sun is d the potential of the two bodies as calculated at a distance x from the Sun is: [latex]-\frac{GM}{x}+\frac{Gm}{d-x}[/latex] The effect is null for .....[latex]x=\frac{Md}{M+m} \approx d(1-\frac{m}{M})[/latex]. This works out to be about 279 mi from the Earth center or 275 mi above its surface. This falls approximately into the Earth exosphere

this is incorrect, guess is a bad thing in physics

There is no such thing as "photon time". Photons do not experience time.

The electromagnetic wave behavior is dictated by the Maxwell equations. The Maxwell equations are invariant wrt the Lorentz transforms. Therefore, your above claim is false. BTW: you have already been told that massive objects (like the rocket) cannot travel at c.

Post 8 is a worked example using your exact data. We don't do the homework for you, you need to put in some effort.

Look at post 8.

This is why you need to go back to your class and see what you were told in terms of calculating [latex]\frac{\partial{\vec{v_i}}}{\partial{\vec{v_j}}}[/latex] when [latex]\vec{v_i}, \vec{v_j}[/latex] are BOTH CONSTANTS.(the link I cited deals with the case when the two vectors are VARIABLE). Do you think you can do that? I am willing to bet that the class notes contain the same exact formulas I came up with. This has nothing to do with the problem you are asked to solve, this is not about "dividing" two matrices.

Because the vectors in his example have CONSTANT components. If you plug in the matrices calculated by the method I suggested, you get the results in the handout. Did you even bother to read the exercise statement? Had you read it, you too could have figured it out. Or maybe not.

First line of the matrix is: [latex]\frac{v_1(1)}{v_{1st}(1)}, \frac{v_1(1)}{v_{1st}(2)},\frac{v_1(1)}{v_{1st}(3)}[/latex] The second line is: [latex]\frac{v_1(2)}{v_{1st}(1)}, \frac{v_1(2)}{v_{1st}(2)},\frac{v_1(2)}{v_{1st}(3)}[/latex] The third line is: [latex]\frac{v_1(3)}{v_{1st}(1)}, \frac{v_1(3)}{v_{1st}(2)},\frac{v_1(3)}{v_{1st}(3)}[/latex]

Yes, it does, except the elements are obtained thru trivial divisions.

What have you tried on your own?

Yes, you can, this is used routinely for calculating series limits.

What you need is the dfinition of the derivative of a vector wrt a vector. The result is a n x n matrix, see here.

It is not "differentiated", formula (16) is just a summation.

You are surely wrong. Why don't you take a break from posting and take a class in vector calculus? There is no such thing as "velocity of image", there is no answer to the nonsense you post.

No, you do not want to learn, what you want to do is to continue posting nonsense. If you wanted to learn, you would have started by taking a break from posting and by taking a class in vector calculus. There is no such thing as [latex]c-n_1[/latex]. Light speed doesn't add/subtract. There is no such thing as [latex]c-(n_1+n_2)[/latex].Light speed doesn't add/subtract.

There are a lot of errors in your post: 1. There is no such thing as "velocity x time". What exists is [latex]\frac{d \vec{r}}{dt}=\vec{v}[/latex] 2. No knowledgeable person claims that "velocity is arc-length over time", you need to stop creating strawmen. 3. "the velocity of light can be any value less than the speed of light" is absolute nonsense since the "value" of the light velocity is ....its speed. And the speed of light is....c. You need to LEARN before you post. So far, you have not learned anything.