Gravitation above and below the earth's surface

In summary, the two points are the maximum and minimum possible distances from the center of the Earth, and the angle between the directions is the maximum and minimum possible angle.
  • #1
erisedk
374
7

Homework Statement


The value of acceleration due to gravity at a point P inside the Earth and at another point Q outside the Earth is g/2 (g being acceleration due to gravity at the surface of earth). Maximum possible distance in terms of radius of Earth R between P and Q is

Ans: R/2 ( 2*root2 + 1 )

Homework Equations


Acceleration due to gravity at a depth d under the Earth = g(1-(d/R))
Acceleration due to gravity at a height h above the Earth = g(1-(2h/R))
g-acc. due to gravity at surface
R- radius of Earth

The Attempt at a Solution


g(1-(d/R))=g(1-(2h/R))
d=2h

g/2 = g(1-(2h/R))
g/2 = g(1-(d/R))
d=R/2

h=d/2=R/4

So, d+h= 3R/4

In all these steps, not once did I think about the maximum possible distance.
And my answer is also wrong.
 
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  • #2
Your expressions for the acceleration of gravity above and below the surface of the Earth are incorrect.

Look up Newton's Law of Gravitation to see how it varies with the radius.

For the gravitation under the surface, consider only the mass inside the sphere below the test point. The outer shell cancels out in that calculation (although showing this is a bit hairy!).
 
  • #3
They're correct. d and h do not indicate the distances from the center of the earth. In case an object is placed at a depth d under the Earth's surface or a height h above the Earth's surface, it's distance from the center of the Earth is R-d and R+h respectively.
 
  • #4
Your expressions are series expansions of the true expressions and assume that h and d are small in relation to the Earth radius. This is no longer true if you want the gravitational acceleration to be half of that on the Earth surface.
 
  • #5
I say again, look up Newton's Law of Gravitation.
 
  • #6
Orodruin said:
Your expressions are series expansions of the true expressions and assume that h and d are small in relation to the Earth radius. This is no longer true if you want the gravitational acceleration to be half of that on the Earth surface.
One of them is correct, though - yes?
 
  • #7
haruspex said:
One of them is correct, though - yes?

Yes, although I will let the OP figure out which of them for him-/herself.
 
  • #8
I got (for above the surface)
g/2 = GM/(R+h)^2
For below the surface,
g/2 = GM/(R-d)^2

So, substituting GM/R^2 for g, I get

2R^2 = (R-d)^2
or R^2 = d^2 - 2Rd
&
R^2 = h^2 + 2Rh

How do I find h+d in terms of R with the pesky +2Rh and -2Rd terms?
 
  • #9
erisedk said:
For below the surface,
g/2 = GM/(R-d)^2

This equation states that gravity would be increasing as you go down, which is not true. This would be true if all of the Earth's mass was collected in a point at the center. This is not the case, but you rather need to use a method that will give you the gravitational constant inside a spherical mass distribution.

Edit: On a side-track. It is probably easier to figure out the radii rather than to figure out the height and depth relative to the Earth surface.
 
  • #10
Oh ok.
M= p(density)*4/3 pi R^3
So, mass that contributes to acceleration due to gravity at a distance r(b) [r(b)<R] from the center of the Earth is (M*r(b)^3)/R^3
So, g/2 = GM r(b) / R^3
&
g/2 = GM / r(a)^2
where r(a) is the distance above the Earth from the center of the earth.

Substituting g as GM/R^2, I get
r(b) = R/2
and r(a) = +R*(root 2) or -R*root 2

Distance between the two points is r(a) - r(b) = R/2 [2*root 2 - 1] or R/2 [2*root 2 + 1]

Since I require the maximum possible distance, the answer is R/2 [2*root 2 + 1].

Thank you so much everybody :D
 
  • #11
Just one remark: Radius is by definition a positive number. However, the directions are free. Your solutions correspond to the max and min of the possible distances, where the directions are opposite/the same. Any distance in between these two are allowed for some angle between the directions.
 

FAQ: Gravitation above and below the earth's surface

1. What is the difference between gravitational force above and below the earth's surface?

The gravitational force above the earth's surface is weaker than it is below the surface. This is due to the inverse square law, which states that the force of gravity decreases as the distance between two objects increases. As you move further away from the center of the earth, the force of gravity decreases. Additionally, the shape of the earth affects the strength of gravitational force, as the earth is not a perfect sphere.

2. How does the force of gravity change as you move further below the earth's surface?

The force of gravity increases as you move further below the earth's surface. This is because there is more mass below you, pulling you towards the center of the earth. The deeper you go, the more mass there is above you, resulting in a stronger gravitational force.

3. Why do objects feel weightless in space?

Objects feel weightless in space because they are in a state of free fall. This means that the only force acting on them is the force of gravity, which is the same as the objects' weight. However, because there is no surface or other forces to counteract this weight, the objects appear to be weightless.

4. How does gravity affect objects in orbit?

Gravity is responsible for keeping objects in orbit around a larger object, such as the earth. The gravitational force between the two objects keeps the smaller object in a constant state of free fall, causing it to continuously circle around the larger object.

5. Can gravitation be observed and measured above and below the earth's surface?

Yes, gravitation can be observed and measured using various instruments such as a gravimeter or accelerometer. These instruments can measure the strength of gravitational force and any changes in it, providing valuable data for scientific research and exploration.

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