Gravitational acceleration

1. Nov 25, 2005

bticu

THIS IS NOT A HOME WORK, THIS IS JUST FOR FUN???

I've been given a challenge by my grade 11 physics teacher and I can't make heads of tail of it, I’ve got the formulas but the numbers don't make sense.
If you went up 109 km (above sea level) and dropped a rock, what speed would it be traveling at and how long would it take to reach 34 km (above sea level). Air resistance can be ignored.
Gravitational constant
=9.79 m/second*second
This is the only formulas he gave me.
Where do I even start???
Distance:
Distance to be traveled = 109km - 34km
Distance to be traveled = 75km
I then started making a chart going by 0.5 second increments
Example:
{Time} (Acceleration) [calulation]
{0.5 sec} (39.16m) [9.79/(0.5*0.5)]
{1 sec} (9.79) [9.79/(1*1)]
{1.5 sec} (4.35) [9.79/(1.5*1.5)]
{2 sec} (2.44) [9.79/(2*2)]
{2.5 sec} (1.57) [9.79/(2.5*2.5)]
{3 sec} (1.09) [9.79/(3*3)]
So could someone tell me is this is right
After 3 seconds, the rocks velocity is 58.4 M/S and traveled a distance of 156 meters
I got the distance by taking the time
{Time} (Acceleration) <Current speed> [calculation]
{0.5 sec} (39.16m) <39.16 m/s> [39.16]
{1 sec} (9.79) <48.95 m/s> [39.16+9.79]
{1.5 sec} (4.35) <53.30 m/s> [48.95+4.35]
{2 sec} (2.44) <55.75 m/s> [53.30+2.44]
{2.5 sec} (1.57) <57.32 m/s> [55.75+1.5]
{3 sec} (1.09) <58.40 m/s> [57.32+1.09]
I got my distance traveled by summing up my current speeds and deviding the total by 2 (2 samples per second speed is in meters per second)
My total speed is the total of my currentspeed
First thing is that correct?
Second thing is this is correct is there a better way of doing this?
If this is not correct, what is the correct calculation?

Last edited: Nov 25, 2005
2. Nov 25, 2005

finchie_88

Just want to say before I begin that this might not be rite, and there is a good chance that the graphics wont come out quite how I want them to, but do your best with it.
I think that I can do part of it, so here goes...
First, we know that the following is true.
$$F_g = \frac{GM_1M_2}{r^2}$$
and
$$F = M_2a$$
(Assuming that mass is constant)
Since air resistance can be ignored, you can work out the velocity from the change in gravitational potential to kinetic energy, so...
$$\frac{1}{2}M_2v^2 = \delta\frac{GM_1M_2}{r}$$
So, rearrange for v and let r be the difference in radius from the centre of mass of the earth, so r includes the radius of the earth.
Now for the time. From the two equations before, we get
$$a = \frac{GM_1}{r^2}$$,
and using calculus, the following can be done...
$$dv/dt = \frac{GM_1}{r^2}$$
So, t can be found by integration (Let one of the limits equal zero) since we know what v is. Note, that the value of the radius is in metres, and takes into account of the radius of the earth, as before.
Hope this is rite, and that it helps.

Last edited by a moderator: Nov 25, 2005
3. Nov 25, 2005

pervect

Staff Emeritus
You might start by looking at

http://www.physicsclassroom.com/Class/1DKin/U1L5d.html

(or start at the beginning with)

http://www.physicsclassroom.com/Class/1DKin/U1L5a.html

If you are given the acceleration of gravity, it's likely your teacher intends you to believe its constant. More advanced calculations are possible which take into account the weakening of gravity as one gets further away from the Earth, but they require calculus to fully justify.

4. Nov 25, 2005

daveed

have you learned the kinematics equations?