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## Homework Statement

At a distance R above the Earth's surface, a particle at rest is let go. Given the gravitational acceleration: [itex] a =-g\frac{R^2}{r^2}[/itex] where

**r**is the distance from the center of the Earth to the particle,

**R**is the Earth radius and

**g**is the gravitational acceleration at the surface, calculate (a) the velocity of the particle when it hits the Earth, (b) the time it takes to fall to the surface. (Earth radius = [itex]R = 6.380 km[/itex], [itex]g=9.81 m/seg^2[/itex]) Results:

**V=7910 m/s, t=34.5 m = 2070 s**

## Homework Equations

1.[itex] a =-g\frac{R^2}{r^2}[/itex]

2.[itex] a = vdv/dr[/itex]

3.[itex] v = dr/dt[/itex]

4.[itex] a = dv/dt[/itex]

## The Attempt at a Solution

Okay, so for (a) what I did was integrating the acceleration using 2. and get velocity as a function of

**r**or the distance Earth center - particle from [itex]r[/itex] to [itex]r_{0}[/itex], these being

**R**to

**2R**, [itex]\int-gR^2/r^2 dr[/itex] and I got [itex]V(r)=\sqrt{2gR^2(\frac{1}{R} -\frac{1}{2R})}[/itex] since [itex]V_{0}[/itex] is 0 because the particle is initially, at

**r=2R**and

**t=0**, at rest. For

**r=R**, it gave me

**7911 m/s**which is the solution the book gives. Now as for the time, I tried integrating the indefinite integral of the acceleration again, to get the time, by using 3.

**dt=dr/v**, which is [itex]\int-gR^2/r^2 dr[/itex] and gives [itex]V(r)=\sqrt{2gR^2/r}+C[/itex] that didn't work, because when I integrate V, I have the constant, and to get that constant I did

**V(r=2R)=0**, because it is at rest, that gave me

**C=-7911**, and that's is impossible because I tried

**V(r=R)**which should be

**7911**, but it gave me

**3277**, so I don't know what I'm doing wrong. I tried the other way, by doing

**V(r=R)=7911**, but the constant didn't correspond to

**V(r=2R)=0**.

The result is

**34.5 m**or

**2070 s**. I also tried integrating the acceleration over time, by using 4., but that got me nowhere, since I would have to use double integrals because

**r**varies. Though I'm not really sure about this last part.