1. ## relative masses

A space traveller is on a mission to determine the mass of the recently discovered
planet U, whose radius is known to be three times that of the earth. In a heroic act the
traveller measures his weight on U to be 130N. If he weighs 700N on earth, what is the
mass of U relative to that of the earth?

2. What ideas have you had so far?

3. Originally Posted by Ackbeet
What ideas have you had so far?
I can work out the gravitational force on planet U as 1.82.
I thought mass was a constant, do they mean weight, in which case why is it on about the radius.

4. Originally Posted by poirot
I can work out the gravitational force on planet U as 1.82.
I thought mass was a constant, do they mean weight, in which case why is it on about the radius.
You can answer your own question by looking at the units of the question. Newtons are units of what?

5. Originally Posted by Ackbeet
You can answer your own question by looking at the units of the question. Newtons are units of what?
force.

6. Right. So that tells you that you will need to take the radii into account, because the gravitational force varies as the inverse square of the distance between the centers of mass of the two attracting bodies. What's the gravitational force law?

7. Originally Posted by Ackbeet
Right. So that tells you that you will need to take the radii into account, because the gravitational force varies as the inverse square of the distance between the centers of mass of the two attracting bodies. What's the gravitational force law?
[LaTeX ERROR: Convert failed] . I've worked out his mass on Earth to be 71.43. I think I fundamentally don't understand this question. I thought W=mg.

8. Technically, the first equation is

$F=-\dfrac{G m_{1}m_{2}}{r^{2}},$

where $G$ is the gravitational constant (not 9.8 meters per second squared). The $W=mg$ equation is a useful approximation when everything is happening at more or less the same distances from the Earth's center. Essentially, you have

$g=\dfrac{G m_{\text{earth}}}{r_{\text{earth}}^{2}}.$

The first equation in this post is the more correct equation, especially when your distance away from Earth's center is changing by a lot.

Does that make sense?

Your computation of the mass is correct. What more can you do?

9. Originally Posted by Ackbeet
Technically, the first equation is

$F=-\dfrac{G m_{1}m_{2}}{r^{2}},$

where $G$ is the gravitational constant (not 9.8 meters per second squared). The $W=mg$ equation is a useful approximation when everything is happening at more or less the same distances from the Earth's center. Essentially, you have

$g=\dfrac{G m_{\text{earth}}}{r_{\text{earth}}^{2}}.$

The first equation in this post is the more correct equation, especially when your distance away from Earth's center is changing by a lot.

Does that make sense?

Your computation of the mass is correct. What more can you do?
what is g

10. g is the usual g in the formula W = mg. It's 9.8 meters per second squared.

11. Originally Posted by Ackbeet
g is the usual g in the formula W = mg. It's 9.8 meters per second squared.
You just said it wasn't. Anyway For the force between the man and Earth I have [LaTeX ERROR: Convert failed] . Do I have to find these out? AS for the force between the man and U [LaTeX ERROR: Convert failed] . There are too many unknowns

12. Originally Posted by poirot
You just said it wasn't.
You're not distinguishing between capital G and lowercase g. Take a look at post # 8 again.

Anyway For the force between the man and Earth I have [LaTeX ERROR: Convert failed] . Do I have to find these out? AS for the force between the man and U [LaTeX ERROR: Convert failed] . There are too many unknowns
The radii need to be squared. I don't think you're going to need to know the radius of the Earth. What you do need to know is that the second planet has a radius that's three times the other.

Try taking the ratio of one equation to the other, and see what cancels out.

13. Originally Posted by Ackbeet
You're not distinguishing between capital G and lowercase g. Take a look at post # 8 again.

The radii need to be squared. I don't think you're going to need to know the radius of the Earth. What you do need to know is that the second planet has a radius that's three times the other.

Try taking the ratio of one equation to the other, and see what cancels out.
sorry you're not really helping. My equations are clearly wrong so that wouldn't help.

14. These are the corrected equations:

$700=\dfrac{71.43 G m_{\text{earth}}}{r_{\text{earth}}^{2}},$ and

$130=\dfrac{71.43 G m_{\text{planet}}}{(3r_{\text{earth}})^{2}}.$

What you're after is an expression of the form

$m_{\text{planet}}=km_{\text{earth}}.$

You can get that by dividing one equation by the other. Now do you follow?

15. Originally Posted by Ackbeet
These are the corrected equations:

$700=\dfrac{71.43 G m_{\text{earth}}}{r_{\text{earth}}^{2}},$ and

$130=\dfrac{71.43 G m_{\text{planet}}}{(3r_{\text{earth}})^{2}}.$

What you're after is an expression of the form

$m_{\text{planet}}=km_{\text{earth}}.$

You can get that by dividing one equation by the other. Now do you follow?
ok I got k=1/9. How do you derive the formula for weight from this equation for F? Thanks a lot for your help.

Also I was a bit confused about the meaning of 'relative'. For example if there were 2 trains going in the same direction, one travelling at 50,the other at 30m/s, you would say the relative speed of the former was 20. But here we are dividing to find the relative mass. Why the difference?

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