What is the velocity of an object dropped from a high altitude?

[zimo]

Homework Statement

An object (mass=1000kg) is dropped from 498,447 metres above the Earth's surface, what will be it's velocity upon impact?

Homework Equations

Energy equivalency.

The Attempt at a Solution

2G*(Earth's Mass)*((1/(earth radius))-(1/earth radius + 498,447))=v^2

v[tex]\approx[/tex]3,001 m/s

Everything looks O.K., but the book has a different answer, am I doing it right and the book is wrong or vice versa?

Thank you.

 

[Niles]

The PE at the top will equal the KE before impact:

m*g*h = ½*m*v^2

I get v = 98,94 m/s


EDIT: And welcome to physicsforums.com :-)

 

[zimo]

Hmmm... sorry for the misunderstanding, but -

First,all of the above numbers in my original message was written in UK/US style - meaning a comma ',' is used as a thousands separator (1,000= a thousand, 1,000,000 = a million and so on). Do you all prefer to use the SI numbering, using a blank space as a separator (1 000 000 for a million, for example)?

Second, I used the universal gravitation equations in much the same way you used the mgh for local-close-altitude equations.

Please, review my solution in the original message and find what's wrong with it, if you may,

Thanks in advance,

UPDATE: I took those parameters as Radius and Mass of the earth: R=6.4*
10^6, M=5.98*10^24)

 

[rootX]

I also got the same answer ><.
What's the answer in the book?

 

[zimo]

I checked it with my tutor, the book is wrong :-)

Viva la studention!

 

[BlackWyvern]

g = GM / r^2
= 8.45492803 N/kg

Ep = mgh
= 1000 . 8.454 . 498,447
= 4,214,333,512 J

Ek = mv^2
4214333512 = 1000.v^2
v = sqrt(4214333.512)
= 2052 ms^-1

That's what I got.

 

[BlackWyvern]

Just having a little bit of a think here, gravity will increase as the object is falling. This increase in gravity can be thought of as an extra acceleration, this means that the force being applied on the mass will vary with position of the mass. I'm not sure how to include this in the calculation.

Ep = mgh, I think only applies when the h is small enough that g doesn't change significantly. So, we'll only get an approximation.

Correct me if I'm wrong.