Relating Gravitational Field Strength and Mass

  • B
  • Thread starter taetae
  • Start date
  • #1
3
0
I know the formula for calculating field strength is g= GM/r2 , however if I'm trying to show the proportionality relationship between just g and m, would g ∝ m be correct, since a larger mass equals a stronger force of gravity and vice versa?
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,600
3,286
Hello tae2, :welcome:

What about the dependence of ##r## on m (c.q. vice versa) ? Think two planets of the same composition ...

PS do use consistent notation: don't change capital M for lower case m if it is not the intention to designate another variable
 
  • #3
3
0
Hello tae2, :welcome:

What about the dependence of ##r## on m (c.q. vice versa) ? Think two planets of the same composition ...

PS do use consistent notation: don't change capital M for lower case m if it is not the intention to designate another variable
Yup, that's where I got confused haha. The mass of the object needs to be divided by the square of it's radius. The question I'm stuck on asks for me to write a proportionality relationship for g relating to m, would g∝ m/r2 be more accurate then? Or should the constant of proportionality, G, be included as well, making it g∝ Gm/r2?

(Taking a class online is so frustrating when you don't quite understand something, I feel like I'm teaching myself!)
 
Last edited:
  • #4
1,521
618
I'm also confused by what your teacher is asking. The relation between mass and gravity is the gravitational constant. It's just a constant proportion. It came about by experimentation. Newton understood the concept that two objects pulled each other and that that force was relative to the distance. He had a problem with that he needed a force of certain units and his equations had a completely different unit, so he added the G and gave it a unit. Later on once we had more precise measurements of mass and force, we were able to discover the value of this constant.

Of course then Einstein came along.
 
  • #5
BvU
Science Advisor
Homework Helper
2019 Award
13,600
3,286
Exercise means gravitational acceleration at the surface, I would venture.

With ##m\propto r^3## and ##g \propto m/r^2## I would say: 8 times heavier, then twice the gravitational acceleration... but I agree that an online answer is a gamble

I feel like I'm teaching myself)
isn't so bad at all ! :smile:
 
  • #6
3
0
Thank you for the help! I hope it's ok to sneak one other question into here (I went through and did all the rest of the questions but this is the only other one I'm confused with). I just want to check whether or not I did it correctly.
It's about determining the orbital period of an asteroid (in earth years) that has the average radius of orbit as 2.77 AU. It says to use the value of Kepler's constant "expressed in the units of yr2/AU3" and the formula T2=Cr3. So, if my understanding is correct, to find the value of Kepler's constant you would do:
C=T2/r3
C=12/13 = 1yr2/AU3

And then to calculate the orbital period:
T=√Cr3
T=√(1yr2/AU3)(2.77)3

I feel like this doesn't look right...I'm pretty sure I didn't input the correct value for C but I'm unsure how to otherwise get it..
 

Related Threads on Relating Gravitational Field Strength and Mass

  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
8
Views
3K
Replies
1
Views
1K
Replies
7
Views
2K
Top