Physics 11 help

  • Thread starter FilthyOtis
  • Start date
  • #1
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Homework Statement



If the force of gravity(weight) of a 1.2kg textbook were to be measured at various distances (d) from the Earth's center, the data would appear as follows:

Force(weight)[kg*m^2/s^2]
1) 11.8
2) 5.21
3) 1.9
4) 1.0

Distance[m]
1) 6.4 x 10^6
2) 9.6 x 10^6
3) 16 x 10^6
4) 22 x 10^6



B) Find the distance, d, at which the gravitational force on the book is 8.0 kg*m/s^2




The Attempt at a Solution



I'm struggling with physics here...

I'm not entirely sure what my equation should be and why. Is it

F(force) = k(constant) / d(distance)^2

?

if so then all I need to do is find k and I can do that correct? but how exactly do I get k.

When I graph the inverse of the square I get a relatively straight line and the table looks like this

Force data stays the same

distance(1/d^2)
1) 24.4 x 10^-15
2) 10.9 x 10^-15
3) 3.9 x 10^-15
4) 2.1 x 10^-15


so.. to find k do I go..

(11.8 - 1.0) / [(24.4 x 10^-15)-(2.1 x 10^-15)]

which on my calculator is 4.843049327 x 10^14 and then plug it into the equation for k and plug in8.0kg*m/s^2 for F and solve for d?

also how do I deal with the units when finding k?

or do I leave out the x 10's and just have 24.4 - 2.1? or am I doing that totally wrong?

How does the weight of the book(1.2kg) come into play here in the equation? or does it?

I'm having troubles understanding why I'm doing what I'm doing in these questions I've been given, which isn't making things easier.

Thank you for any help you can provide, I have another question as well but am not sure if I should post 2 separate posts or combine the 2 different questions in the same post?

- Otis
 

Answers and Replies

  • #2
cristo
Staff Emeritus
Science Advisor
8,122
74

Homework Statement



If the force of gravity(weight) of a 1.2kg textbook were to be measured at various distances (d) from the Earth's center, the data would appear as follows:

Force(weight)[kg*m^2/s^2]
1) 11.8
2) 5.21
3) 1.9
4) 1.0

Distance[m]
1) 6.4 x 10^6
2) 9.6 x 10^6
3) 16 x 10^6
4) 22 x 10^6



B) Find the distance, d, at which the gravitational force on the book is 8.0 kg*m/s^2




The Attempt at a Solution



I'm struggling with physics here...

I'm not entirely sure what my equation should be and why. Is it

F(force) = k(constant) / d(distance)^2

?
Yes, this is correct.

if so then all I need to do is find k and I can do that correct? but how exactly do I get k.

When I graph the inverse of the square I get a relatively straight line and the table looks like this
Ok, so have you plotted the graph F vs. d-2?

Force data stays the same

distance(1/d^2)
1) 24.4 x 10^-15
2) 10.9 x 10^-15
3) 3.9 x 10^-15
4) 2.1 x 10^-15


so.. to find k do I go..

(11.8 - 1.0) / [(24.4 x 10^-15)-(2.1 x 10^-15)]
If the graph is what I have guessed, then this is the correct way of finding the gradient (which is k in this case.)
which on my calculator is 4.843049327 x 10^14 and then plug it into the equation for k and plug in8.0kg*m/s^2 for F and solve for d?
yup
also how do I deal with the units when finding k?
I'm not sure what you mean by this.
or do I leave out the x 10's and just have 24.4 - 2.1? or am I doing that totally wrong?
No, you were correct above.

How does the weight of the book(1.2kg) come into play here in the equation? or does it?
The mass(!) of the book comes into play in the constant k. Since it is the same every time, we just put this in the value of k. If you are interested, the general law for the gravitational force is [tex]F=\frac{Gm_1m_2}{r^2}[/tex] where G is called the gravitational constant and m_1 and m_2 will be masses of the book and the earth.

Thank you for any help you can provide, I have another question as well but am not sure if I should post 2 separate posts or combine the 2 different questions in the same post?

- Otis

It depends whether the questions are related or not. Either way, you should probably try and finish one problem before attempting another.
 
  • #3
15
0
Thank you, I ended up figuring it all out before seeing this but I appreciate the reply!
 

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