# Basic Newtonian Questions #2

by PhysicsNovice
Tags: basic, newtonian
 P: 4 6a 7b 8b 9c 10b I think you are not familiar with the g-r graph. Try studying it and you'll find it helpful in solving a lot of problems.
P: 9
 6. If the radius of earth somehow decreased with no change in mass, your weight would: a. increase b. not change* c. decrease
We know that Weight of a body on Earth (W) = mass of the body (m) x acceleration due to gravity (g).

But $$F=\frac{GM}{R^2}$$ where M is the mass of the Earth and R the radius of the Earth.

Combining the two equations, we see that m is inversely proportional to $$R^2$$.

I will put up the rest of the answers next time 'coz I really have to run now.

Bye!!!

P: 46

## Basic Newtonian Questions #2

 P: 46 Thanks Deydas. It is these equations that make my head spin. But I think I get the logic. I responded to John54 expaning problem 6. Am I correct that the answer should be (a) increase for the reasons that I stated. Your weight is dependant on the mass of the planet you are on and the distance you are from the center of the planet? If the planet's mass stays constant but the radius gets smaller your weight has to increase. I hope this is the correct explanation. Also, in your equation is the F = gravitational force and G some constant? Also, how did you merge the two equations to show that your weight is inversely proportional to the radius squared? Thanks for assisting a true beginning physicist (I say this with tongue in cheek).
Mentor
P: 40,883
 Quote by PhysicsNovice I will start with #6. If the radius of earth somehow decreased with no change in mass, your weight would: (a.) increase, (b.) not change, (c.) decrease. I am changing my answer to (a) increase. My reason is as follows (my words trying to explain the equation F=GM divided by R squared): The gravitational force responsible my weight is equal to some constant G x the Earths mass divided by the Earth's radius squared. So if we keep the mass of the Earth constant and shrink the Earth's radius then the resulting gravitational force would be greater because the radius squared is much smaller.
This is correct.

To be clearer: Your weight equals the gravitational attraction between you and the earth. That force equals:
$$F = G M m / R^2$$, where M is the mass of the earth and m is your mass. In problem #6 the only thing that changes is that R gets smaller--thus your weight increases.
Mentor
P: 40,883
 Quote by deydas We know that Weight of a body on Earth (W) = mass of the body (m) x acceleration due to gravity (g).
This is true.
 But $$F=\frac{GM}{R^2}$$ where M is the mass of the Earth and R the radius of the Earth.
I assume you mean $g = GM/R^2$.

 Combining the two equations, we see that m is inversely proportional to $$R^2$$.
I assume you mean that a body's weight is inversely proportional to $R^2$. This is true.
 P: 46 Hello Doc Al. I knew it would not be long to hear from you when you saw me struggling. Will I ever have a "physicists" way of thinking? I hope at least to be able to capture the basics. Thanks for you assistance again. I know that I am too logical, sequential, anal-retentive but these formulas just mess up my thinking when I can not see the connections. I get the concepts but sometimes the formulas make me re-think or change my mind in processing the information. Even in your explanation I am wondering if the g is equal to the F and in your statement "Combining the two equations, we see that m is inversely proportional to R squared. I assume you mean that a body's weight is inversely proportional to R squared." The m is not even in the equation unless you mean it is included in the F because it is equal to GMm/R squared. Again, please say that I am correct. Thanks for your patience.
 P: 46 John54. Hello again. I research #7 The force of gravity acting on you will increase if you (a.) burrow deep inside the planet, (b.) stand on a planet with a radius that is shrinking, (c.) both of these, or (d.) none of these. I am not sure why the answer is not what I originally posted (c.) both of these. I thought that the weight of an object equals the gravitational attaction between the two objects (person and planet) and that the body weight is inversely proportional to the planet's radius squared. Or if mass of the planet increses or it diameter (radius) gets smaller the resulting force (weight) would increase. If you dig a hole in the planet and get closer to the planets center (same as decreasing the radius) this would result in an increased weight. Where did I go wrong?
Mentor
P: 40,883
 Quote by PhysicsNovice Even in your explanation I am wondering if the g is equal to the F and in your statement "Combining the two equations, we see that m is inversely proportional to R squared. I assume you mean that a body's weight is inversely proportional to R squared." The m is not even in the equation unless you mean it is included in the F because it is equal to GMm/R squared. Again, please say that I am correct. Thanks for your patience.
Right. Note that "combining the two equations" was not my statement, but that of deydas. I was merely correcting it.

I think deydas was trying to say this:
(1) W = mg
(2) g = GM/R^2

Thus W = GMm/R^2. The weight is inversely proportional to the square of the Earth's radius.

That's fine, but you could have started with that answer based on Newton's law of gravity.
Mentor
P: 40,883
 Quote by PhysicsNovice If you dig a hole in the planet and get closer to the planets center (same as decreasing the radius) this would result in an increased weight. Where did I go wrong?
Digging a hole into the planet is not the same as merely decreasing the radius. In problem #6 the mass of the planet remained constant while the radius shrank, but that's not true in problem #7.

As you tunnel into the planet, only the mass underneath you affects your weight (assuming a symmetrical distribution of mass). Assuming a uniform density, the mass underneath you is proportional to the radius cubed. That ends up making the net gravitational force on you--your weight--directly proportional to the radius. So your theoretical weight decreases from a maximum at the surface to zero at the center.
 P: 46 Doc Al. Sorry about posting in two forums. I did not know that was a rule. Also, was not sure if General or Classical was the way to go. I know my questions are very basic but I enjoy the site. I think that I am learning as much here as I am with my on-line learning system and readings. Thanks for your patience.
 P: 9 physicsnovice, you may try out Fundamentals of Physics by Resnick, Halliday and Walker. The text is wonderful and the concepts (along with the formulae) is very clearly explained. And I am sorry for not being able to explain the problems properly. I will try my best next time. Thank you.

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