What is the Gravitational Acceleration at a Height Equal to Earth's Diameter?

In summary: What is the symbol for density?The density of a substance is the mass of a unit of the substance per unit of volume.
  • #1
Deebu R
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Homework Statement


If the gravitational accleration at the Earths surface is 9.81 m/s^2 what is its value at a height equal to the diameter of the Earth from its surface?

2. The attempt at a solution
I have heard that it becomes 1/4th the value as the center of mass is moving away form each other.
But I really can't understand the idea. Why is it 1/4?
 
Last edited:
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  • #2
What do you know about the force gravity?
 
  • #3
PeroK said:
What do you know about the force gravity?
I know that it is a force acting between 2 masses with a distence between them. It is represented by F= G m1m2/ d^2.
I also know that it is a concervative force and a central force.
 
  • #4
Deebu R said:
I know that it is a force acting between 2 masses with a distence between them. It is represented by F= G m1m2/ d^2.
I also know that it is a concervative force and a central force.

That's good. So, what can you do with that formula for ##F##?
 
  • #5
PeroK said:
That's good. So, what can you do with that formula for ##F##?
Well I don't understand how I can apply it here. How can we use the formula of force to find gravitational accelereration? Why not find the gravitational acceleration at a hight h from surface of Earth?
 
  • #6
Deebu R said:
Well I don't understand how I can apply it here. How can we use the formula of force to find gravitational accelereration? Why not find the gravitational acceleration at a hight h from surface of Earth?

Are force and acceleration not related?
 
  • #7
PeroK said:
Are force and acceleration not related?
Yes but I still don't understand how that solve the problem... Say I use the formula G=9.81m/s^2. Earths radius is 6400km so its diameter is x2. what about mass?
should it be volume x density? Won't that make the problem unnecessarily large and complicated?
 
  • #8
Deebu R said:
Yes but I still don't understand how that solve the problem... Say I use the formula G=9.81m/s^2. Earths radius is 6400km so its diameter is x2. what about mass?
should it be volume x density? Won't that make the problem unnecessarily large and complicated?

One thing at a time! Let me help. You have a general formual for the gravitational force. You want the gravitational acceleration due to the Earth:

1) On the Earth's surface

2) At a height equal to the diameter of the Earth above its surface.

So, you need:

Let ##M## be the mass of the Earth, let ##R## be the radius of the Earth. Now, what can you can about the accleration of gravity at the Earth's surface?
 
  • #9
PeroK said:
One thing at a time! Let me help. You have a general formual for the gravitational force. You want the gravitational acceleration due to the Earth:

1) On the Earth's surface

2) At a height equal to the diameter of the Earth above its surface.

So, you need:

Let ##M## be the mass of the Earth, let ##R## be the radius of the Earth. Now, what can you can about the accleration of gravity at the Earth's surface?
Acceleration due to gravity at the Earth's surface = GM/R^2
 
  • #10
Deebu R said:
Acceleration due to gravity at the Earth's surface = GM/R^2

What about above the Earth's surface, at a height equal to the Earth's diameter?
 
  • #11
PeroK said:
What about above the Earth's surface, at a height equal to the Earth's diameter?
Well, using Gp= GM/(R+h)^2. So...9.81 x M/ (6400+12800)^2 ?
Isn't M= volume x density. How can I find M withought density?
 
  • #12
Work with symbols. Leave it in as M. It's the same for h = 0 as for any other value of h
 
  • #13
So... 9.81 x M /(6400)^2+h^2?
 
  • #14
No. And I still see numbers, so I haven't been clear enough.
You have
1) at ground level $$g = {GM\over R^2}$$
2) at height h $$g^* = {GM\over (R+h)^2}$$All you want to do is pick a good value for h and express ##g^*## in terms of ##g##
 
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  • #15
BvU said:
No. And I still see numbers, so I haven't been clear enough.
You have
1) at ground level $$g = {GM\over R^2}$$
2) at height h $$g^* = {GM\over (R+h)^2}$$All you want to do is pick a good value for h and express ##g^*## in terms of ##g##
Already pointed out both equations but I don't get the value part...Earth's radius=6400km so shouldn't its diameter be 6400 x 2.
Is it the convertion? should I make it into meter?
 
  • #16
Also what is the value of M?
 
  • #17
Ahhhh I see G is the gravitational constent 6.674x 10^11 and M = gr^2/G.Right?I though G was g.This was my own mistake.
But still can't get the right answer. Maybe I need to understand the concept more deeply.
 
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  • #18
In BvU's post # 14, algebraically, what is g*/g equal to? Do you need to know the mass of the Earth M to get this ratio?
 
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  • #19
Chestermiller said:
In BvU's post # 14, algebraically, what is g*/g equal to? Do you need to know the mass of the Earth M to get this ratio?
Okay. I have to find the ratio..so GM cancel out...

so g*/g= R^2/(R+h)?
 
  • #20
Deebu R said:
Okay. I have to find the ratio..so GM cancel out...

so g*/g= R^2/(R+h)?
Yes, except the R+h is also squared. Does this give you your desired answer?
 
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  • #21
Deebu R said:
Okay. I have to find the ratio..so GM cancel out...

so g*/g= R^2/(R+h)?

You just have to figure out the relationship between ##R## and ##h## now.
 
  • #22
Chestermiller said:
Yes, except the R+h is also squared. Does this give you your desired answer?
(6400)^2 /( 6400+12800)^2= 0.111?

Also I don't understand why I found the ratio and why was Earth's acceleration given if we never use it?
 
  • #23
Deebu R said:
(6400)^2 /( 6400+12800)^2= 0.111?

Also I don't understand why I found the ratio and why was Earth's acceleration given if we never use it?

You haven't used it yet! You just need to think another step ahead.
 
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  • #24
PeroK said:
You haven't used it yet! You just need to think another step ahead.
I don't get it
 
  • #25
Deebu R said:
I don't get it

Read the question!
 
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  • #26
so g=9.81m/s^2?
g* is gravitation at height h

we found the ratio of g* and g =0.11

Are you saying that g*= 0.11 x 9.81?
 
  • #27
Deebu R said:
so g=9.81m/s^2?
g* is gravitation at height h

we found the ratio of g* and g =0.11

Are you saying that g*= 0.11 x 9.81?

I'm not saying that! But, if you're saying that, then you're right.
 
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  • #28
PeroK said:
I'm not saying that! But, if you're saying that, then you're right.
g*=0.11 x 9.81=1.079m/s^2.
Is that the answer?
 
  • #29
This is actually a question from a previous year question paper.
Available options are:
a) 4.905m/s^2 b)2.452 m/s^2 c)3.27 m/s^2 d) 1.09 m/s^2
 
  • #30
Deebu R said:
This is actually a question from a previous year question paper.
Available options are:
a) 4.905m/s^2 b)2.452 m/s^2 c)3.27 m/s^2 d) 1.09 m/s^2

It can only be one of those!
 
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  • #31
Deebu R said:
g*=0.11 x 9.81=1.079m/s^2.
Is that the answer?
What if I told you that 0.11111...= 1/9

What would your answer be then?
 
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  • #32
PeroK said:
It can only be one of those!
Well yeah. I guess...what will I do if the options are like that?Should I go for option d?
 
  • #33
Chestermiller said:
What if I told you that 0.11111...= 1/9

What would your answer be then?
Actually I got the answer as 1/9 but since the option given are in decimal form I divided it to get 0.1111...
 
  • #34
Deebu R said:
Actually I got the answer as 1/9 but since the option given are in decimal form I divided it to get 0.1111...
Well, 9.81/9 = 1.09
 
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  • #35
Chestermiller said:
Well, 9.81/9 = 1.09
OHHHHHHH! I did not think of that. This is mind blowing. I am happy. Thank you very much for your time and effort. It felt really really nice.
 

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