What happens to the gravitational field strength's magnitude if

AI Thread Summary
When the radius (r) decreases by a factor of 4, the gravitational field strength's magnitude increases by 16 times due to the inverse square law, which states that gravitational force is inversely proportional to the square of the distance. Conversely, if r increases by a factor of 2, the gravitational field strength's magnitude decreases to one-fourth of its original value. The discussion clarifies that the correct interpretation involves changing r by a factor rather than subtracting from it. This principle applies broadly to other inverse square laws, such as those governing light and sound intensity. Understanding these relationships is crucial for solving related physics problems effectively.
chaishreen
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Homework Statement


What happens to the gravitational field strength's magnitude if
a) r decreases by a factor of 4?
b) r increases by factor of 2

Homework Equations


i'm not sure which equation they are referring to? it may be g ∝ 1/r^2

The Attempt at a Solution


I tried to do this:
1/ (r-4)^2 and i got 1/(r^2 - 8r + 16)
the answer at the back of the book for a) says gravitational field strength's magnitude is supposed to become 16 times greater and for b) it's supposed to become 1/4 as great
please help
 
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chaishreen said:

Homework Statement


What happens to the gravitational field strength's magnitude if
a) r decreases by a factor of 4?
b) r increases by factor of 2

Homework Equations


i'm not sure which equation they are referring to? it may be g ∝ 1/r^2


The Attempt at a Solution


I tried to do this:
1/ (r-4)^2 and i got 1/(r^2 - 8r + 16)
the answer at the back of the book for a) says gravitational field strength's magnitude is supposed to become 16 times greater and for b) it's supposed to become 1/4 as great
please help

The radius has not decreased by 4, but by a factor of 4.

So r → r/4 (rather than r → r-4)
 
hey thanks, but is it g ∝ 1/(r/4)^2?
 
Fg=Gmm/r^2 so if you bring the masses 4 times closer and Gmm remain the same, then Fg' = Gmm/(r/4)^2. It follows that Fg'=Gmm/(r^2/4^2) which is Gmm/r^2/16, thus 16 (Gmm/r^2). Since Gmm/r^2is the original force, the new one is 16 times greater.

Likewise you can show that increasing r to 2r will make (2r)^2 = 4r^2, so the force will be 4 times smaller.
 
chaishreen said:
hey thanks, but is it g ∝ 1/(r/4)^2?

Expand that expression and what do you get? Compare it to the original 1/r2
 
chaishreen said:

Homework Statement


What happens to the gravitational field strength's magnitude if
a) r decreases by a factor of 4?
b) r increases by factor of 2

Homework Equations


i'm not sure which equation they are referring to? it may be g ∝ 1/r^2


The Attempt at a Solution


I tried to do this:
1/ (r-4)^2 and i got 1/(r^2 - 8r + 16)
the answer at the back of the book for a) says gravitational field strength's magnitude is supposed to become 16 times greater and for b) it's supposed to become 1/4 as great
please help

Also a simple approach is: this is an example if an "inverse square law" [some others are intensity of light, intensity of sound, electrical attraction between charges ...]

The inverse part tells you the change is opposite: reduce the separation - increase the force.
The square part tells you the size of change. change "r" by a factor of 4 → a "g" change of 42 (which is 16).
 
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