# Why is my speed imaginary?

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?

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Why don't you just try the conservation of energy. mdvdt=mvdvdR is false. – Shaktyai Sep 21 '12 at 16:35
In the first line, shouldn't the force point in the direction of decreasing R? Shouldn't the force have a negative sign? – Alfred Centauri Sep 21 '12 at 16:37
A free body diagram would have helped here. I think $v$ is measured in the opposite sense as $R$ causing a problem. – ja72 Sep 21 '12 at 17:52

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

$v_e = \sqrt{\dfrac{2GM}{R}}$

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Awesome you are the best! – Cactus BAMF Sep 21 '12 at 17:51
@CactusBAMF, thanks and one final note: the distance from the surface of the planet is $r_0-R$ in my notation so if your problem gives the initial distance, make sure to add the planet's radius to it to get $r_0$ in my formula. – Alfred Centauri Sep 21 '12 at 18:56