|
A particle is released from a dense atmosphere free planet of radius r at a distance R (R >> r ). find the speed on impact. $$F = \frac{GmM}{(r + R)^2} = m \frac{dv}{dt} = mv\frac{dv}{dR}$$ $$GM\int^{R_f}_{R_0}(r + R)^{-2} dr = \int^{V_f}_{V_o}vdv$$ But $R_f$ = 0 when the object strikes the planet $$-GM[r^{-1} - (r +R_0)^{-1}] = 1/2[v_f^2 - v_0^2]$$ $$-2GM[\frac{R_0}{r(r + R_0)}] + V_0^2 = V_f^2$$ why is it imaginary? |
|||||||||||
|
|
First, I think the essential problem is that the gravitational force points in the direction of decreasing distance so the force formula should have a negative sign. Also, your notation is mixed up. You should be integrating with respect to the radial coordinate $R$, not the constant $r$. But, it would be more conventional to denote the constant radius of the planet with $R$ and the radial coordinate with $r$. Assume that convention in the following: $F = -\dfrac{GmM}{r^2} = m \dfrac{dv}{dt} = m \dfrac{dv}{dr} \dfrac{dr}{dt} = m v \dfrac{dv}{dr}$ Integrating both sides with respect to the radial coordinate: $-GM \int^{R}_{r_0}r^{-2} dr = \int^{v_R}_{v_0}vdv$ $GM[\dfrac{1}{R} - \dfrac{1}{r_0}] = 1/2[v_R^2 - v_0^2]$ $2GM[\dfrac{1}{R} - \dfrac{1}{r_0}] + v_0^2 = v_R^2$ For $r_0 = \infty$ and $v_0 = 0$, we recover the escape velocity formula $v_e = \sqrt{\dfrac{2GM}{R}}$ |
|||||
|
