Q. Resisted Motion (1 Viewer)

[ezzy85]

from patel:

An object is projected vertically upwards with the initial velocity U from the earths surface. The accelaration obeys the law given by:

d²x/dt² = -k/x²

where x is the distance of the particle from the centre of the earth whose is R.
Given the accelaration is g when x = R, show that velocity v in any position is given by:

v² = U² - 2gR²(1/R - 1/x)

Hence show that U = 12kms^-1 for the escape velocity.

I need help with this hole topic really...

 

[spice girl]

use da/dt = vdv/dx

initially we have velocity = U, displacement at ground altitude (i.e. displacement x = R)

so we have I{U->v} vdv = I{R->x} -k/x^2 dx

this should get you (1/2)(v^2 - u^2) = k(1/x - 1/R)

Given the accelaration is g when x = R => da/dt = g = k/R^2
thus k = gR^2

so v^2 = u^2 - 2k(1/R - 1/x) = u^2 - 2gR^2(1/R - 1/x)

escape velocity is when x -> infinity, v = 0 (when projectile JUST manages to escape earth's gravity)
so u^2 = 2gR^2(1/R) = 2gR

plug in data, you may get sqrt(2gR) = 12km/s

 

[Affinity]

Alternative solution without integration:

(1) find k.

by your given conditions we have -g = -k/R^2

k = gR^2

(2)show that E=[(v^2)/2 - k/x ] for the particle is a constant.

dE/dt

= [v * dv/dt ]- [ (-k/x^2)* dx/dt ]

= [v*a - a*v] // a is the acceleration.

= 0

E must be constant since it's derivative with respect to time is 0.

(3) Find E By substituting the initial values.

E=[(U^2)/2 - k/R ]

(4) We have E=[(v^2)/2 - k/x ] holds for any x

so [(U^2)/2 - k/R ] = [(v^2)/2 - k/x ] for any x

(U^2)/2 = (v^2/2) + k/R - k/x

U^2 = v^2 + 2k(1/R - 1/x)

U^2 = v^2 + 2(gR^2)(1/R - 1/x) As required.

(5) find the escape velocity.

v^2 > 0 at all times for particle to escape.
x-> infinity for particle to escape.

U^2 > 0 + 2(gR^2)(1/R - 0)
U^2 > 2gR
U> sqrt(2gR)

and I presume they give you numbers for g and R at this stage to substitute and evalute.

If you do physics:

I choosed the value E=[(v^2)/2 - k/x ] to work with because it is the total energy of the particle divided by it's mass. that is (KE + GPE)/m

for uniform acceleration E could be chosen to be (v^2)/2 + gx where g is the magnitude of the acceleration.