Another gravitation problem

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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0

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.
 
<h2>1. What is another gravitation problem?</h2><p>Another gravitation problem refers to a scientific question or phenomenon that involves the force of gravitation, which is the attraction between two objects with mass. This could include problems related to the motion of planets, the orbit of satellites, or the behavior of objects in outer space.</p><h2>2. How is gravitation measured?</h2><p>Gravitation is typically measured using the universal law of gravitation, which states that the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This can be expressed mathematically as F = G(m1m2)/r^2, where F is the force of attraction, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.</p><h2>3. What is the difference between gravitation and gravity?</h2><p>Gravitation and gravity are often used interchangeably, but there is a subtle difference between the two terms. Gravitation refers to the force of attraction between two objects with mass, while gravity refers to the overall effect of this force on the objects. In other words, gravity is the result of gravitation.</p><h2>4. How does gravitation affect the motion of objects?</h2><p>Gravitation plays a major role in the motion of objects in space. It is responsible for keeping planets in orbit around the sun, moons in orbit around planets, and satellites in orbit around Earth. Gravitation also affects the trajectory of objects, causing them to accelerate towards each other due to the force of attraction.</p><h2>5. What are some real-life applications of gravitation?</h2><p>Gravitation has many practical applications in our daily lives. It is used in the design and operation of satellites, which are crucial for communication, navigation, and weather forecasting. Gravitation also plays a role in the formation of celestial bodies, such as stars and galaxies. Additionally, understanding gravitation allows us to accurately predict and study astronomical events, such as eclipses and planetary alignments.</p>

1. What is another gravitation problem?

Another gravitation problem refers to a scientific question or phenomenon that involves the force of gravitation, which is the attraction between two objects with mass. This could include problems related to the motion of planets, the orbit of satellites, or the behavior of objects in outer space.

2. How is gravitation measured?

Gravitation is typically measured using the universal law of gravitation, which states that the force of attraction between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This can be expressed mathematically as F = G(m1m2)/r^2, where F is the force of attraction, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

3. What is the difference between gravitation and gravity?

Gravitation and gravity are often used interchangeably, but there is a subtle difference between the two terms. Gravitation refers to the force of attraction between two objects with mass, while gravity refers to the overall effect of this force on the objects. In other words, gravity is the result of gravitation.

4. How does gravitation affect the motion of objects?

Gravitation plays a major role in the motion of objects in space. It is responsible for keeping planets in orbit around the sun, moons in orbit around planets, and satellites in orbit around Earth. Gravitation also affects the trajectory of objects, causing them to accelerate towards each other due to the force of attraction.

5. What are some real-life applications of gravitation?

Gravitation has many practical applications in our daily lives. It is used in the design and operation of satellites, which are crucial for communication, navigation, and weather forecasting. Gravitation also plays a role in the formation of celestial bodies, such as stars and galaxies. Additionally, understanding gravitation allows us to accurately predict and study astronomical events, such as eclipses and planetary alignments.

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