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

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Homework Help Overview

The discussion revolves around determining the gravitational acceleration at a height equal to the Earth's diameter from its surface, starting from the known value of 9.81 m/s² at the surface. Participants explore the implications of gravitational force and its relationship with distance from the Earth's center.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formula for gravitational force and its application to find gravitational acceleration at different heights. There are questions about the relationship between force and acceleration, and how to express gravitational acceleration in terms of known values.

Discussion Status

Several participants are engaged in exploring the mathematical relationships and ratios involved in gravitational acceleration. Some guidance has been provided on how to express gravitational acceleration at height in relation to its value at the surface, but there remains some confusion about the calculations and the use of mass in the equations.

Contextual Notes

Participants are working under the constraints of a homework problem, which includes specific values for Earth's radius and the requirement to express gravitational acceleration at a height equal to the Earth's diameter. There is also a mention of available multiple-choice answers that may influence the discussion.

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Homework Statement


If the gravitational acceleration 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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What do you know about the force gravity?
 
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.
 
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##?
 
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 height h from surface of Earth?
 
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 height h from surface of Earth?

Are force and acceleration not related?
 
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?
 
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 acceleration of gravity at the Earth's surface?
 
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 acceleration 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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