Why does the gravity of a shrunken Earth increase by nearly 2%?

[Deebu R]

Homework Statement

If the radius of of the Earth were to shrink by one percent, its mass remains the same,g would decrease by nearly...%

Homework Equations

g=GM/ R^2

The Attempt at a Solution

Well, I took g1= GM/R^2

And gravity after shrinking as g2= GM/(R-(R/100))^2
Then I did g1-g2 and I didn't get the answer. [/B]

 

[Bystander]

Start with a calculation of the new volume.

 

[Deebu R]

V1= 4/3 pi r^3

V2= 4/3 pi (r-r/100)^3

 

[Bystander]

would decrease by nearly...%

Next, is this the exact statement?

 

[Deebu R]

Next, is this the exact statement?

Yes, doesn't gravitation decrease to zero when we go to the core?

 

[Bystander]

Yes, but we're talking about surface gravity.

 

[Deebu R]

Yes, but we're talking about surface gravity.

I understand but if Earth shrinks then isn't it like moving towards the core?
Also the question does say decrease it was not an error.

 

[jbriggs444]

Start with a calculation of the new volume.

What does volume have to do with it when density is not remaining constant? Post #1 already has the correct formula.

I understand but if Earth shrinks then isn't it like moving towards the core?

As you move toward the core, you dig down, leaving a portion of the Earth above your head. The portion of the Earth that is farther from the core than you are ceases to have any net gravitational effect on you.

If the Earth shrinks beneath your feet, the entire mass of the Earth remains beneath your feet and continues to affect you.

 

[Bystander]

Where is the mass located relative to the new surface, interior, or exterior?

 

[Deebu R]

What does volume have to do with it when density is not remaining constant? Post #1 already has the correct formula.

As you move toward the core, you dig down, leaving a portion of the Earth above your head. The portion of the Earth that is farther from the core than you are ceases to have any net gravitational effect on you.

If the Earth shrinks beneath your feet, the entire mass of the Earth remains beneath your feet and continues to affect you.

So if the mass remains the same gravity does not change?

 

[jbriggs444]

So if the mass remains the same gravity does not change?

If you consider Newton's universal law of gravity:

$F = \frac{G m_1 m_2}{r^2}$

indeed gravity does not change. But if you want to keep your feet on the surface of the shrunken Earth than, as Bystander suggests, it is surface gravity that you need to consider. Your original post correctly considered the effect of this in reducing the r term.

Edit: The problem statement asks for a percentage decrease. What percentage decrease did you compute?

 

[Deebu R]

If you consider Newton's universal law of gravity:

$F = \frac{G m_1 m_2}{r^2}$

indeed gravity does not change. But if you want to keep your feet on the surface of the shrunken Earth than, as Bystander suggests, it is surface gravity that you need to consider. Your original post correctly considered the effect of this in reducing the r term.

Edit: The problem statement asks for a percentage decrease. What percentage decrease did you compute?

I got g1 - g2 =98GM/99

How can I convert this to percentage?

Also if you think about it in terms of

g is inversely proportional to R^2 then shouldn't gravity increase as radius decrease? So won't that make the question wrong?

 

[jbriggs444]

I got g1 - g2 =98GM/99

I do not understand that answer at all. Where did the 98 come from, where did the 99 come from and what are the units on that answer?

Edit: I have a suspicion that there is a rounding error, but the R went away as well, so it's more egregious than that.

g is inversely proportional to R^2 then shouldn't gravity increase as radius decrease? So won't that make the question wrong?

It makes the question "interesting".

 

[Deebu R]

g1= GM/R^2

g2= GM/ (R-R/100)^2.

Using (a-b)^2 rule

g2= GM/ 100R^2-2R^2+R^2 = GM/99R^2

g1- g2 = GM/R^2 - GM/99R^2

 

[Bystander]

This could be a "re-tread" problem; it "expands" one year, "contracts" the next, and the subsequent parts aren't always kept "in sync" with the rest of the statement.

 

[Deebu R]

This could be a "re-tread" problem; it "expands" one year, "contracts" the next, and the subsequent parts aren't always kept "in sync" with the rest of the statement.

So the question is wrong?

 

[Deebu R]

It makes the question "interesting".

Well I guess you can say it's interesting when it poses no threat. But imagine sitting in an exam hall and this shows up and you hear a voice in your head saying "your God, where is he now?"...

 

[Bystander]

No. It is out of synchronism with itself; it has to be interpreted carefully.

 

[jbriggs444]

g1= GM/R^2

g2= GM/ (R-R/100)^2.

Using (a-b)^2 rule

g2= GM/ 100R^2-2R^2+R^2 = GM/99R^2

That is incorrect. Please learn to use parentheses and decimal points.

A correct result is $g_2 = \frac{GM}{(0.99R)^2}$

 

[Deebu R]

Is the right answer 1 percent?

 

[gneill]

Is the right answer 1 percent?

Helpers won't directly confirm or deny an answer unless you show exactly how you arrived at it.

 

[Deebu R]

I did g1- g2 this time g2= GM/0.99R^2 and got GM-0.99GM/ 0.99R^2 = 0.01/0.99

Then I divided it with the original value GM/R^2 and got (1/99) x 100= 1.01

So I thought 1 percent

 

[jbriggs444]

I did g1- g2 this time g2= GM/0.99R^2 and got GM-0.99GM/ 0.99R^2 = 0.01/0.99

Try again, M O R E S L O W L Y

Use parentheses and justify every step.

 

[Deebu R]

Taking g1-g2 is the correct method?

 

[jbriggs444]

Taking g1-g2 is the correct method?

Sure. But you need to do it right without muffing the algebra. Your problem is not the physics. It's the algebra.

 

[Deebu R]

g2= ((100GM)/(99R^2))^2 = (100/99)^2 x g1 [since g1= GM/R^2]
= 1.02 g1
g1-g2= g1-1.02(g1)= -0.02g1

[(g1-g2)/g1] x 100= -0.02 x 100= -2%

Since the percentage is negative I am guessing g is actually increasing not decreasing. Then

[(g2-g1)/ g1] x 100 = +2%

In that case I believe the question was actually wrong since -2% was not one of the MCQ's. Also +2% is a provided choice.