Physics and non-constant g

I was thinking last night that all the calculations for mechanics and physics use a constant gravitational field strength. How would one go about calculating the kinetic energy of a projectile, or max height (you know normal projectile questions) with varying g. I know f=GMm/r^2 is involved but not sure how. I was thinking i could integrate this equation which would give the area under a f against r graph, right, which is the same as force*distance, which is work done -> energy. But i think im making lots of mistake here, not even sure if this is the way to go about it. Any help?

Reply 1

corkskrew

The kinetic energy is the energy at a given point in time

E = 0.5*m*v*v

So it takes into account the mass of the moving body and the velocity at which it is moving.

In relation to a varying g, g is a vector and gives the acceleration due to gravity at a given point from the centre of a mass.

I think what you might be getting at is a problem whereby, for example:
A mass of 2000kg is moving at 10ms-2 away from the surface of the earth. How high will it get.

The equation V= - GM/r should be used here where V is the potential energy at a given point. It's a negative value as it is defined as the energy required toraise a mass from infinity to that distance (r) from the centre of a mass (M)
To answer it, the mass which is moving from the earth's surface has E=0.5*m*v*v which is equal to 100000J

It therefore has 100000J to "spare" (neglect air resistance and all that malarkey)

V=-GM/r

V = the energy at the earth's surface
G = gravitational constant
M = mass of the central object, in this case the earth's mass = 6x10^24kg
r = distance from the centre of the object (at earth's surface, = earth's radius = 6.4x10^6m)

So at the surface, energy = GM/r = 62531250J
And the energy being lost in this direction is the kinetic energy, = 100000J
So final energy = 62531250J - 100000J = 62431250J

From the same equation, the final distance away from the centre is found to be:

V = -GM/r

62431250 = -(G x 6x10^24) / r

So, r = (G x M)/62431250
r = 6410251m

This distance is from the centre of the earth, so distancemoved = this distance minus the "centre to surface" distance, equal to

d = 6410251 - 6400000
d = 10251m

Reply 2

Ipsen

And when the initial velocity is not radial, remember to resolve it in to the transverse component a, which is constant as the gravitational field is radial, and the radial component b. At the maximum height, b=0. Sub this into the energy equation and off you go :smile:

Reply 3

riddles55

wow, fantastic explaination corkskrew and ipsen, thanks a lot! Good luck with the exams!

Reply 4

riddles55

I've just tested this out a couple of times but it doesnt seem to work...

I'm comparing newtonian particle projectile method (uvast) to the V=-GM/R, to see the error between the two.

Firstly, let u = 20m/s directly upwards,

u=20, a = -9.8, v=0 (i'm finding max height)

v^2 - u^2 = 2as
20^2 = 2*9.8*s
s = 20.408m (5 sig figs)
This is roughly how high the next method should give, just a little bit less.

Ok, V= -GM/R
Values from google, google "mass of the earth" to see what i mean.
G=6.673EXP-11
Mass=5.9742EXP24
R=6.378EXP6

using these values, V=62477131 (tonnes of sig figs so i dont lose accuracy, just in case)

Kinetic energy of a 1kg particle at 20m/s= 0.5mv^2 = 0.5*400 = 200J

62477131 - 200 = 62476931, let this equal V

62476931 = GM/R

R = 6381423

6381423- radius of earth = height of projectile
=> 6381423-6378000 = 3423m

Much higher than the 20m previously calculated...