# Integration and forces

• struggles
In summary, the conversation discusses the use of Newton's law and the equation f = −gmx/R for an accurate description of motion in a deep hole. The discussion then focuses on deriving an equation for the speed and position of a parcel dropped into a hole that goes through the center of the Earth. The process involves using x ̈ = vdv/dx and integrating with appropriate limits. The result involves trigonometric functions, with a final equation of x = Rsin(t√g/√R). However, there may be a mistake with the signs, as the next part of the question suggests that the result should involve cos instead of sin.

## Homework Statement

The question is pretty long and wordy so apologies in advance!

Inside the Earth the gravitational field falls off linearly as one approaches the centre. An accurate description of motion in a very deep hole would therefore be to use Newton’s law, f = ma, but with the force being

f = −gmx/R

where m is the object’s mass, x is the distance of the object from the centre of the Earth, R is the radius of the Earth, and g is the usual acceleration due to gravity measured at the surface.

1. (i) Suppose that a parcel destined for Australia is dropped from rest at ground level into a hole that goes through the centre of the Earth. Derive an equation for the speed of the subsequent motion v(x) (where v = x ̇). [Hint: Use x ̈ = vdv/dx.]
2. (iii) Find an expression for the parcel’s position, x(t).
3. (iv) Using the θ ≪ 1 approximation, cosθ ≈ 1 − θ2 + ..., show that when the displacement is small, (R − x) ≪ R, your expression for x(t) gives the usual result, x ≈ R − gt2/2 .
4. (v) Show (using the approximations for R and g in (ii)) that it takes about an hour for the parcel to arrive in Australia.

## The Attempt at a Solution

1) I tried -gmx/R = mx.. = mvdv/dx
I then separated the variables and integrated
-gm/R∫xdx = mvdv with the limits of x from 0-->2R and the limits for v 0-->v
this gave me -4gmR = mv2
Rearranging for v (I'm not sure about the sign) v = 2√gr

2) Assuming the above equation is correct (which I'm fairly cercain it isn't!) I then simply integrated with respect to time to get x = 2t√gr. This is obviously wrong as it doesn't work in the next part parts.

I'd really appreciate any help, thanks!

struggles said:

## Homework Statement

The question is pretty long and wordy so apologies in advance!

Inside the Earth the gravitational field falls off linearly as one approaches the centre. An accurate description of motion in a very deep hole would therefore be to use Newton’s law, f = ma, but with the force being

f = −gmx/R

where m is the object’s mass, x is the distance of the object from the centre of the Earth, R is the radius of the Earth, and g is the usual acceleration due to gravity measured at the surface.
1. (i) Suppose that a parcel destined for Australia is dropped from rest at ground level into a hole that goes through the centre of the Earth. Derive an equation for the speed of the subsequent motion v(x) (where v = x ̇). [Hint: Use x ̈ = vdv/dx.]
2. (iii) Find an expression for the parcel’s position, x(t).
3. (iv) Using the θ ≪ 1 approximation, cosθ ≈ 1 − θ2 + ..., show that when the displacement is small, (R − x) ≪ R, your expression for x(t) gives the usual result, x ≈ R − gt2/2 .
4. (v) Show (using the approximations for R and g in (ii)) that it takes about an hour for the parcel to arrive in Australia.

## The Attempt at a Solution

1) I tried -gmx/R = mx.. = mvdv/dx
I then separated the variables and integrated
-gm/R∫xdx = mvdv with the limits of x from 0-->2R and the limits for v 0-->v
Those limits of integration are incorrect.

The parcel goes from the surface of the earth, looks like x = R, the way you have thing set up.

The parcel starts at rest.

Then you want to find v at some arbitrary value of x. So the x integral goes from R to x. The v integral goes from 0 to v.
this gave me -4gmR = mv2
Rearranging for v (I'm not sure about the sign) v = 2√gr

2) Assuming the above equation is correct (which I'm fairly cercain it isn't!) I then simply integrated with respect to time to get x = 2t√gr. This is obviously wrong as it doesn't work in the next part parts.

I'd really appreciate any help, thanks!

Ok i still can't get this to work out:
so after integrating i get -gm/R(x2/2 -R2/2) = mv2/2
Rearranging i get v = √(gR - gx2/R) = dx/dt

Then rearranging to integrate again -
dx(gR - gx2/R)-1/2 = 1dt
and integrating
R/gx (gR - gx2/R)1/2

This isn't right but I've played around and can't get it to a reasonable answer

struggles said:
Ok i still can't get this to work out:
so after integrating i get -gm/R(x2/2 -R2/2) = mv2/2
Rearranging i get v = √(gR - gx2/R) = dx/dt

Then rearranging to integrate again -
dx(gR - gx2/R)-1/2 = 1dt
and integrating
R/gx (gR - gx2/R)1/2

This isn't right but I've played around and can't get it to a reasonable answer
Check that last integration.
The derivative of your result does not give the integrand.

You might try a trig substitution. Part (3) gives a clue that trig functions are involved.

SammyS said:
Check that last integration.
The derivative of your result does not give the integrand.

You might try a trig substitution. Part (3) gives a clue that trig functions are involved.

so eventually after making substitutions i get √r/√g ∫ du/√(1-u2) = √r/√g arcsin(x/r) (where u = x/r)
and rearranging get x = Rsin(t√g/√R). However the next part of the question ( and that at t=0 c should = r) implies that it should be cos. Any ideas or have i just made a slip with signs somewhere along the way (potentially the order in which i applied the limits in the first integration?) Thanks so much!

struggles said:
so eventually after making substitutions i get √r/√g ∫ du/√(1-u2) = √r/√g arcsin(x/r) (where u = x/r)
and rearranging get x = Rsin(t√g/√R). However the next part of the question ( and that at t=0 c should = r) implies that it should be cos. Any ideas or have i just made a slip with signs somewhere along the way (potentially the order in which i applied the limits in the first integration?) Thanks so much!
What are the limits for that integration?

Also, I assume you have used r and R interchangeably.

SammyS said:
What are the limits for that integration?

Also, I assume you have used r and R interchangeably.

So the limits would be t is from 0 to t and x from R to x which gives √R2/g2(π/2 - arcsin(x/r)) = t.
Rearranging sin(π/2 - t√(g2/R2)) = cos(t√(g2/R2) = x! Thank you! think I've finally got there!

struggles said:

## Homework Statement

The question is pretty long and wordy so apologies in advance!

Inside the Earth the gravitational field falls off linearly as one approaches the centre. An accurate description of motion in a very deep hole would therefore be to use Newton’s law, f = ma, but with the force being

f = −gmx/R

where m is the object’s mass, x is the distance of the object from the centre of the Earth, R is the radius of the Earth, and g is the usual acceleration due to gravity measured at the surface.

1. (i) Suppose that a parcel destined for Australia is dropped from rest at ground level into a hole that goes through the centre of the Earth. Derive an equation for the speed of the subsequent motion v(x) (where v = x ̇). [Hint: Use x ̈ = vdv/dx.]
2. (iii) Find an expression for the parcel’s position, x(t).
3. (iv) Using the θ ≪ 1 approximation, cosθ ≈ 1 − θ2 + ..., show that when the displacement is small, (R − x) ≪ R, your expression for x(t) gives the usual result, x ≈ R − gt2/2 .
4. (v) Show (using the approximations for R and g in (ii)) that it takes about an hour for the parcel to arrive in Australia.

## The Attempt at a Solution

1) I tried -gmx/R = mx.. = mvdv/dx
I then separated the variables and integrated
-gm/R∫xdx = mvdv with the limits of x from 0-->2R and the limits for v 0-->v
this gave me -4gmR = mv2
Rearranging for v (I'm not sure about the sign) v = 2√gr

2) Assuming the above equation is correct (which I'm fairly cercain it isn't!) I then simply integrated with respect to time to get x = 2t√gr. This is obviously wrong as it doesn't work in the next part parts.

I'd really appreciate any help, thanks!

The differential equation for ##x(t)## is just that of simple harmonic motion, like that of a mass attached to a spring obeying Hook's law. Here, ##x(t)## is displacement from the Earth's center, and can be < 0 as well as > 0.

## 1. What is integration in science?

Integration in science refers to the process of combining different pieces of knowledge or disciplines to form a more complete understanding of a particular phenomenon or concept. It involves synthesizing information from various sources, such as different scientific fields or research studies, to create a holistic view of a topic.

## 2. How does integration contribute to scientific progress?

Integration allows scientists to see connections and patterns between seemingly unrelated ideas or data, leading to new insights and discoveries. It also helps to bridge gaps between different fields of study and promotes cross-disciplinary collaboration, which can accelerate scientific progress.

## 3. What are some examples of forces in science?

Forces are interactions between objects that cause changes in motion or shape. Some common examples of forces in science include gravity, friction, electricity, and magnetism. Other examples include tension, compression, and buoyancy.

## 4. How does the study of forces help us understand the natural world?

The study of forces is essential to understanding how objects move and interact with each other in the natural world. It helps us explain phenomena such as planetary orbits, weather patterns, and the behavior of atoms and molecules. Understanding forces also allows us to develop technologies and tools that improve our everyday lives.

## 5. What are some real-world applications of integration and forces?

Integration and forces have numerous real-world applications in fields such as engineering, physics, biology, and environmental science. For example, engineers use integration to design structures that can withstand different forces, while physicists use it to develop theories and models to explain natural phenomena. In biology, the integration of genetics and evolutionary theory has led to a better understanding of how species evolve. Additionally, the study of forces has applications in developing sustainable energy sources and predicting natural disasters.