Conservation of Energy: Solving for Δv, Integrals, Derivatives

[redrum419_7]

1) [itex]\frac{1}{2}[/itex]mv[itex]^{2}_{2}[/itex]-mgy[itex]_{2}[/itex] = [itex]\frac{1}{2}[/itex]mv[itex]^{2}_{1}[/itex]-mgy[itex]_{1}[/itex]

2) [itex]\frac{1}{2}[/itex]m(v[itex]^{2}_{2}[/itex]-2gy[itex]_{2}[/itex]) = [itex]\frac{1}{2}[/itex]m(v[itex]^{2}_{1}[/itex]-2gy[itex]_{1}[/itex])

since g = [itex]\frac{GM}{R^{2}}[/itex] and [itex]\frac{1}{2}[/itex]m cancels.

3) v[itex]^{2}_{2}[/itex]-2[itex]\frac{GM}{R^{2}}[/itex]y[itex]_{2}[/itex] = v[itex]^{2}_{1}[/itex]-2[itex]\frac{GM}{R^{2}}[/itex]y[itex]_{1}[/itex]

4) v[itex]^{2}_{2}[/itex]-v[itex]^{2}_{1}[/itex] = 2[itex]\frac{GM}{R^{2}}[/itex]y[itex]_{2}[/itex]-2[itex]\frac{GM}{R^{2}}[/itex]y[itex]_{1}[/itex]

5) v[itex]^{2}_{2}[/itex]-v[itex]^{2}_{1}[/itex] = 2[itex]\frac{GM}{R^{2}}[/itex](y[itex]_{2}[/itex]-y[itex]_{1}[/itex])

6) [itex]\frac{Δ(v^{2})}{Δy}[/itex] = 2[itex]\frac{GM}{R^{2}}[/itex], if R is along y-axis

then d(v[itex]^{2}[/itex]) = 2[itex]\frac{GM}{y^{2}}[/itex]dy

Can someone give me a tip on where to go from here? Would an integral or derivative have any significance? Or are there any errors? Any feedback would be greatly appreciated. ( I know v = √2gy )

 

[Doc Al]

1) [itex]\frac{1}{2}[/itex]mv[itex]^{2}_{2}[/itex]-mgy[itex]_{2}[/itex] = [itex]\frac{1}{2}[/itex]mv[itex]^{2}_{1}[/itex]-mgy[itex]_{1}[/itex]

Why the minus signs?

2) [itex]\frac{1}{2}[/itex]m(v[itex]^{2}_{2}[/itex]-2gy[itex]_{2}[/itex]) = [itex]\frac{1}{2}[/itex]m(v[itex]^{2}_{1}[/itex]-2gy[itex]_{1}[/itex])

since g = [itex]\frac{GM}{R^{2}}[/itex] and [itex]\frac{1}{2}[/itex]m cancels.

Note that PE = mgy only holds close to the Earth's surface. g is a constant and R here equals the radius of the earth. If you want to describe PE for larger ranges of distance, you cannot use PE = mgy. For a more general expression for potential energy see: Gravitational Potential Energy

 

[vin300]

Well, doc, you might remember that I was banned for seven days for being stubborn that the potential energy is the work done to bring an object from infinity to the point? The same is said in the link you provided.

 

[Doc Al]

Well, doc, you might remember that I was banned for seven days for being stubborn that the potential energy is the work done to bring an object from infinity to the point? The same is said in the link you provided.

There's nothing wrong with such a statement, when applied to situations where the force goes to zero at infinity.

 

[redrum419_7]

Why the minus signs?

[/URL]

Isn't U(x) = -G(x)

 

[redrum419_7]

from the last part d(v[itex]^{2}[/itex]) = [itex]\frac{2GM}{y^{2}}[/itex]dy, if you take the integral, ∫d(v[itex]^{2}[/itex]) = ∫2[itex]\frac{GM}{y^{2}}[/itex]dy

v[itex]^{2}[/itex] = -[itex]\frac{2GM}{y}[/itex], set v to c and rearrange y, you get

y=-[itex]\frac{2GM}{c^{2}}[/itex]

or

R[itex]_{s}[/itex]=-[itex]\frac{2GM}{c^{2}}[/itex], would this be a valid derivation of the Schwarzschild radius?

 

[Doc Al]

Isn't U(x) = -G(x)

I don't understand your notation.

U(x) = -∫F(x)dx

 

[redrum419_7]

I don't understand your notation.

U(x) = -∫F(x)dx

Yeah G(x) =∫F(x)dx, that's what my teacher uses.

 

[redrum419_7]

I see why it is not a minus sign, because from the integral and and the minus sign, it will be positive.