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kirills
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I've been trying to tackle this question for a while now, but I'm afraid it's not going anywhere without some outside help.
So let's say we have some body falling towards a planet without any atmosphere, and let's assume that the initial velocity is not equal to zero. Given these conditions I'm trying to derive a formula which would allow me to determine the exact time of impact or, in fact, time to any given point along the trajectory of the falling body .
Let's define r0 as initial separation; r as a destination point on the trajectory of the falling body(surface, for example); v0 as initial velocity; v as instantaneous velocity at point r; t0 as initial time; and T as time at point r.
Now acceleration as a function of distance is given by:
[tex]\frac{d^2r}{dt^2}=\frac{GM}{r^2}[/tex]
Using the chain rule we arrive at:
[tex]a(r)=\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}=v\frac{dv}{dr}[/tex] Therefore:[tex]a(r)dr=vdv[/tex]
Equivalently:
[tex]\frac{GM}{r^2}=vdv[/tex]
Now in order to get v:
[tex]v=\frac{dr}{dt}=\sqrt{\frac{2G(m1+m2)}{r}−\frac{2G(m1+m2)+v0^2}{r0}}[/tex]
[tex]v=\frac{dr}{dt}=\sqrt{\frac{2Gr0(m1+m2)−2Gr(m1+m2)+v0^2r0r}{r r0}}[/tex]
dt is calculated as:
[tex]dt=\sqrt{\frac{r0 rdr} {2Gr0(m1+m2)−2Gr(m1+m2)+(v0^2rr0)}}[/tex]
Now I have absolutely no idea how to calculate dr, but I hear it has something to do with gravitational potential energy. Some guidance on this matter would be much appreciated. I also have some doubts regarding the next step of my strategy. Let's say I manage to find dr, and what then? Intuitively I feel this is a right path to take but in fact I have no idea how to convert dr and dt into T.
So let's say we have some body falling towards a planet without any atmosphere, and let's assume that the initial velocity is not equal to zero. Given these conditions I'm trying to derive a formula which would allow me to determine the exact time of impact or, in fact, time to any given point along the trajectory of the falling body .
Let's define r0 as initial separation; r as a destination point on the trajectory of the falling body(surface, for example); v0 as initial velocity; v as instantaneous velocity at point r; t0 as initial time; and T as time at point r.
Now acceleration as a function of distance is given by:
[tex]\frac{d^2r}{dt^2}=\frac{GM}{r^2}[/tex]
Using the chain rule we arrive at:
[tex]a(r)=\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}=v\frac{dv}{dr}[/tex] Therefore:[tex]a(r)dr=vdv[/tex]
Equivalently:
[tex]\frac{GM}{r^2}=vdv[/tex]
Now in order to get v:
[tex]v=\frac{dr}{dt}=\sqrt{\frac{2G(m1+m2)}{r}−\frac{2G(m1+m2)+v0^2}{r0}}[/tex]
[tex]v=\frac{dr}{dt}=\sqrt{\frac{2Gr0(m1+m2)−2Gr(m1+m2)+v0^2r0r}{r r0}}[/tex]
dt is calculated as:
[tex]dt=\sqrt{\frac{r0 rdr} {2Gr0(m1+m2)−2Gr(m1+m2)+(v0^2rr0)}}[/tex]
Now I have absolutely no idea how to calculate dr, but I hear it has something to do with gravitational potential energy. Some guidance on this matter would be much appreciated. I also have some doubts regarding the next step of my strategy. Let's say I manage to find dr, and what then? Intuitively I feel this is a right path to take but in fact I have no idea how to convert dr and dt into T.
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