Height for 1/100 normal weight above Earth's surface

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How high above the surface of the Earth should a rocket be
in order to have 1/100 of its normal weight? Express your answer
in units of Earth radii.

Im not sure were to start with this one. I know that the moon is 60 times as far away as the core of Earth to the surface.
 
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Use the gravitational force equation

[tex]F = G\frac{Mm}{r^2}[/tex]

and

F = mg

Setting the two equations equal to each other provides the acceleration due to gravity as a function of the distance, r, from the center of the earth.
 
You'll need this formula

[tex]F=\frac{GMm}{r^2}[/tex]


Where M is the mass of the Earth and m is the mass of the object.
 
how do you get G or calculate the mass of the earth?And which one of you is right? Yor saying two diffrent things.
 
Mass of the Earth and G are found in any physics book; usually in an appendix.
 
i found the mass of Earth 5.9742*10to the 24 power and the radius but how do i know the mass of the rocket? Or G?
 
The mass of the rocket, m, is not required. When setting F = mg equal to the gravitational force equation the m on each side cancels.

G is the universal gravitational constant. You should be able to find the value in the physics book or on line easily.
 
You already have the answers you need, G is a well known physical constant, get it from a textbook or find it online. And as for the mass of the rocket, equate the equations given by chrisk and you will discover why it is irrelevent.
 
1 6.67300 × 10-11 m3 kg-1 s-2 i hope i made it look right. So what do i put in for Kg and s and m?
 
The appropiate units will cancel. Make sure the units for the Earth radius is in meters.
 
i got 24485606.14 that's not right is it?
 
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I have not done the calculation but express the value in Earth radii then subtract one Earth radius from this value to give the height above the Earth surface.
 
Thats not right, and you still haven't expressed it in units of Earth radii.
 
i looked up the real answer from a website and it said the answer is 57,402 and 9 Earth unit radii.
 
That is the correct answer, if you still don't understand try posting your workings and I'll show you where you have gone wrong.
 
first i did F= G*M/r2
M=5.9742*1024
R=6378.1
G=6.67300 × 10-11
 
after equating the two force equations you should have been left with:

g=G*M/r^2

you need to rearrange this for r and solve it
 
What do you mean equating the two force equations?
 
chrisk said:
Use the gravitational force equation

[tex]F = G\frac{Mm}{r^2}[/tex]

and

F = mg

Setting the two equations equal to each other provides the acceleration due to gravity as a function of the distance, r, from the center of the earth.

those are the force equations!
 
equating:

F = F

LaTeX Code: mg = G\\frac{Mm}{r^2}
 
oops that didnt work, I mean:

F = F
mg = GMm/[r][/2]