# Homework Help: Simple gravity crash question

1. Aug 14, 2007

### zimo

1. The problem statement, all variables and given/known data
An object (mass=1000kg) is dropped from 498,447 metres above the earth's surface, what will be it's velocity upon impact?

2. Relevant equations
Energy equivalency.

3. The attempt at a solution
2G*(Earth's Mass)*((1/(earth radius))-(1/earth radius + 498,447))=v^2

v$$\approx$$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.

2. Aug 14, 2007

### 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 :-)

3. Aug 14, 2007

### 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)

Last edited: Aug 14, 2007
4. Aug 14, 2007

### rootX

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

5. Aug 19, 2007

### zimo

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

Viva la studention!

6. Aug 21, 2007

### 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.

7. Aug 21, 2007

### 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.

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