Mechanics Question. (1 Viewer)

[Aerath]

A body is projected vertically upwards from he Earth's Surface with velocity U. The acceleration of a particle in space is given by k/x^2 towards cenre of the earth, where x is the distance of the body from the centre of the earth. Given that the acceleration is g at the earth's surface, prove that the velocity v, at time t is given by v^2 = U^2 - 2gR^2 ( 1/R - 1/x), where R is the radius of the earth.
(Done that bit)

If U^2 = 2gR, find dx/dt in terms of x and show that the body will reach a distance 8R from the earth's surface, in a time of 2.72 hours.

I keep getting 0.86 =\.

 

[untouchablecuz]

A body is projected vertically upwards from he Earth's Surface with velocity U. The acceleration of a particle in space is given by k/x^2 towards cenre of the earth, where x is the distance of the body from the centre of the earth. Given that the acceleration is g at the earth's surface, prove that the velocity v, at time t is given by v^2 = U^2 - 2gR^2 ( 1/R - 1/x), where R is the radius of the earth.
(Done that bit)

If U^2 = 2gR, find dx/dt in terms of x and show that the body will reach a distance 8R from the earth's surface, in a time of 2.72 hours.

I keep getting 0.86 =\.

what i did was:

sub U^2 = 2gR

simplifies the expression for v^2 considerably

square root both sides and we have an expression for dx/dt

take the reciprical of both sides

integrate with respect to x

to evaluate the constant, when t = 0, x = R

then, to prove what is given

let x = 8R + R = 9R (remember, from the centre of the Earth not the surface)

simplify, then i let g = 10, and R = 6400 km = 6 400 000 m

and i got 9805 s = 2.723... hours = 2.72 hours

i dunno if i did i right (having to use values for g and R)

 

[jet]

I got the same as untouchablecuz. It's probably just rounding errors from the values of g and R. It's close enough anyway.