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

[Deebu R]

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?

 

[PeroK]

What do you know about the force gravity?

 

[Deebu R]

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.

 

[PeroK]

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 $F$?

 

[Deebu R]

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?

 

[PeroK]

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?

 

[Deebu R]

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?

 

[PeroK]

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?

 

[Deebu R]

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

 

[PeroK]

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?

 

[Deebu R]

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?

 

[BvU]

Work with symbols. Leave it in as M. It's the same for h = 0 as for any other value of h

 

[Deebu R]

So... 9.81 x M /(6400)^2+h^2?

 

[BvU]

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 $g$

 

[Deebu R]

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?

 

[Deebu R]

Also what is the value of M?

 

[Deebu R]

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.

 

[Chestermiller]

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?

 

[Deebu R]

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)?

 

[Chestermiller]

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?

 

[PeroK]

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.

 

[Deebu R]

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?

 

[PeroK]

(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.

 

[Deebu R]

You haven't used it yet! You just need to think another step ahead.

I don't get it

 

[PeroK]

I don't get it

Read the question!

 

[Deebu R]

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?

 

[PeroK]

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.

 

[Deebu R]

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?

 

[Deebu R]

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

 

[PeroK]

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!