If falling object already has some speed to center of big mass, then what will be energy change with change of hight?

The total energy is conserved. In the case you're talking about, there are only two relevant types of energy: kinetic and gravitational potential. Their sum must be constant, so changing the potential necessarily changes the kinetic energy. This translates to the equation:

 

[math]\frac{1}{2} m v_1^2 - \frac{GMm}{r_1} = \frac{1}{2} m v_2^2 - \frac{GMm}{r_2}[/math]

 

Edit: I just noticed this is in the relativity section, so if you're looking for the relativistic case then it's a bit different. I'll post the relativistic version of this later, but I'm not at home right now and I'm typing on my phone.

Edited October 4, 201411 yr by elfmotat

The total energy is conserved. In the case you're talking about, there are only two relevant types of energy: kinetic and gravitational potential. Their sum must be constant, so changing the potential necessarily changes the kinetic energy. This translates to the equation:

 

[math]\frac{1}{2} m v_1^2 - \frac{GMm}{r_1} = \frac{1}{2} m v_2^2 - \frac{GMm}{r_2}[/math]

 

Edit: I just noticed this is in the relativity section, so if you're looking for the relativistic case then it's a bit different. I'll post the relativistic version of this later, but I'm not at home right now and I'm typing on my phone.

Thank you.Yes. The formula is useful for me,but the relativistic case could be more useful.

In the relativistic case, you'll have some spacetime metric [math]g_{\mu \nu}[/math] and Christoffel symbols [math]\Gamma^{\mu}_{\alpha \beta}[/math] constructed from the metric. From these you can use the geodesic equation to get a differential equation for the energy of a small particle with respect to some parameter s:

 

[math]\frac{d E(s)}{ds}=- \frac{1}{m} \Gamma^{0}_{\alpha \beta} \, p^{\alpha} (s) p^{\beta} (s)[/math]

 

If you can solve that equation for E(s), then you'll know the energy of the particle at any value of s. If the metric is independent of time (which it usually is), then conservation of energy holds and dE/ds=0.

 

An alternative approach would be to use [math]g_{\mu \nu} p^\mu p^\nu = const.[/math] when the metric is time-independent, where [math]p^0[/math] is the particle's energy.

What is wrong when I use such equation?:

 

(gamma₁ -1)mc² - GMm/r₁ = (gamma₂-1)mc² - GMm/r₂

 

Well, I should add increasing of kinetic energy in increased gravitation

 

(gamma₁ -1)mc² - GMm/r₁ = (gamma₂-1)mc² - GMm/r₂ - (gamma₁ -1)mc² {(1-r₀/r₂)^1/2 -(1-r₀/r₁)^1/2}

 

r_{0 is} Schwarzschild radius

 

Is this correct now?

Edited October 5, 201411 yr by DimaMazin

What is wrong when I use such equation?:

 

(gamma₁ -1)mc² - GMm/r₁ = (gamma₂-1)mc² - GMm/r₂

When gravitational fields are very strong (like around a black hole) that equation doesn't work. The relativistic one I posted is accurate for all macroscopic objects regardless of their speed or the strength of the gravitational field.

Edited October 5, 201411 yr by elfmotat