I have do this:
As I don't know if f is an eigenfunction of Ω, I expand it as ## f = \sum_n c_nf_n## where ##f_n## are orthonormal eigenfunctions of Ω, so:
##\left<Ω\right> = \left<f\left|Ω\right|f\right> = \left<f\left|1*Ω*1\right|f\right> ##
##\sum_m\sum_n...
Homework Statement
Show that if ##\left( \Omega f\right) ^* = -\Omega f ^* ##
then ##\left< \Omega \right> = 0 ## for any real function f. where ##\Omega## is an operator
Homework Equations
It's a self test of the completeness relation --Molecular quantum mechanics (Atkins)--
so the equation...
Finally I got it!
To simplify my expression I try the two ways you said me.
First way:
Δr = \frac {4πr^3f}{Gm_sM_e + 4πfr^2}
Δr = \frac {\frac {4πr^3f}{r^2}}{\frac{Gm_sM_e}{r^2} + \frac {4πfr^2}{r^2}}
Δr = \frac {4πrf}{\frac{Gm_sM_e}{r^2} + 4πf}
Δr = \frac {4πrf}{GF + 4πf}
and considering...
I have corrected my expression and now is dimensionally correct. The units is metres, the meters the radius has decrease in that lap when the radius is r. Now I try to find dr/dt
2πrf = \frac {-Gm_sM_e} {2r} + \frac {Gm_sM_e} {2(r - Δr}
2r(r - Δr) 2πrf = - (r-Δr) Gm_sM_e + rGm_sM_e...
Checking the dimensions of the two expressions, now I see that the book's one is speed units, and mine is dimensionally wrong, so I try to correct it but also try to find the book solution. Thanks!
Homework Statement
A satellite of mass m_s orbits the Earth in a circular orbit of radius r_0. If the satellite orbits at the upper part of the atmosphere and the friction force f is constant, it would trace an spiral and fall to the earth. but if we suppouse that the friction force is small...
Yes, the KE is always positive.
The GPE could be zero at the lowest level and positive at the initial position or begin with zero and negative later. But the SPE is positive because the spring is stretching and gaining energy.
I don't know if you are looking after this, where the kinetic energy appears by calculation of the work done by a force:
$$w= \int_A^B F dx
=\int_A^B m \frac {dv} {dt} dx
\, = m\int_A^B dv \frac {dx} {dt}
=m\int_A^B vdv$$
$$w= m\left[ \frac {v^2} 2 \right] _A^B
= m \frac 1 2 v_B^2 -...
I have the proof for the 8)
As the second derivative of P is too long to do it, It's easier to proof that slope of the curve p(x) with x < b/2 is negative and the slope of the curve
with x > b/2 is positive. I have do it and there is a minimum in x = b/2