Tides Caused by the Moon?

[CabinetMaker]

TEST TIME! (on gravity. )

Imagine an experiment in two parts.

Part 1: drop a spherical, 10kg object from 1km altitude above a perfectly spherical 10²⁴kg planet. Record time from release to impact.

Part 2: Repeat experiment with all the same parameters except for the mass that is to be dropped, which is replaced with a 10¹⁵kg object of equal volume.

Both parts are performed in a vacuum.

Question. How would the times for each part compare?
A: Part 2 would give a shorter time.
B: Part 1 would give a shorter time.
C: Both times would be the same.

C (Using classical mechanics)

 

[Stripe]

C (Using classical mechanics)

Wrong (using classical mechanics).

 

[CabinetMaker]

Wrong (using classical mechanics).

Well, lets take a look and see: Here are the equations that govern you set up.

First up, gravity:
Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them:[3]

where:

• F is the force between the masses;
• G is the gravitational constant (6.673×10−11 N · (m/kg)2);
• m1 is the first mass;
• m2 is the second mass;
• r is the distance between the centers of the masses.

This describes the force of attraction between the two bodies of masses m1 and m2 separated by a distance r. Note that this is a force which is defined as mass times acceleration. Solve the equation for force, set it equal to m*a (F=m*a) and solve for a: (a=F/m)

For case 1 m1=10²⁴ and m2=10. Since you did not provide a radius of the bodies we will use r=1 as you did provide the stipulation that the volume of the two masses is the same. Thus F1=6.73E-11*(1E24*10)/1 = 6.73E14. Since F=ma, a=F/m. In this case, m is the total mass of the systems so a=6.73E14/(1E24+10)=6.73E-10

For case 2 m1=1024 and m2=10²⁴. Since you did not provide a radius of the bodies we will use r=1. Thus F1=6.73E-11*(1E24*1E15)/1 = 6.73E28. Since F=ma, a=F/m. In this case, m is the total mass of the systems so a=6.73E28/(1E24+1E15)=6.73E4.

The acceleration force in case 2 is 1*10¹⁴ times larger than in case 1.


Next, you need to solve for position which is given by this equation: x=a*t^2+V*t+c where a is acceleration, v is the initial velocity (zero in this case) and c is the initial position. The equation reduces to: x=a*t^2+1. At t=0, x=1. We want to solve for t when x=0. t=sqrt((x-1/a)). Since the acceleration for case 2 is much greater than the acceleration for case 1, case 2 will have a much shorter time.

The correct answer is A. My bad. In the equation for gravitational force I made the error of confusing G with g.

 

[Psalmist]

The tides, as explained by Newton and accepted by majority opinion, are caused by the gravity of the moon. This thread isn't meant to explain Newton's position. It is assumed this position is known by the reader. The point is to raise questions concerning this theory.

1. If the earth's gravity is stronger than that of the moon, how does the moon's gravity over power that of the earth regarding the tides of the ocean?

2. Why is it not thought that the sun's gravity causes the tides?

3. Why aren't other objects affected by the moon's gravity?

4. If the moon's gravity is uniform, why are the tides not uniform?

5. Why aren't ALL bodies of water affected in this way, such as lakes, ponds and puddles?

How can the moon's gravity function in this manner?

6. Slack Tide, don't forget slack tide.​

 

[Psalmist]

6. Slack Tide, don't forget slack tide.​

Slack Tide
Before any turn of the tide, there is a time of slack water or slack tide.
The occurrence of relatively still water at the turn of the (low) tide.
​

 

[Stripe]

The correct answer is A.

:thumb:

Can you explain why (in layman's terms)?

 

[gcthomas]

:thumb:

Can you explain why (in layman's terms)?

The planet itself will accelerate more and move a greater distance when the falling body is more massive, so the falling body does not have to fall as far.

 

[Stripe]

The planet itself will accelerate more and move a greater distance when the falling body is more massive, so the falling body does not have to fall as far.

:thumb:

 

[Stripe]

So — ignoring air resistance — to the old question of which falls faster, a brick or a feather, we should say the brick. Even if the difference is utterly negligible (exponentially more so if dropped together).

Right?

 

[CabinetMaker]

:thumb:

Can you explain why (in layman's terms)?

Let me give it a go.

Every object attracts every other objects to itself. The heavier the object is, the more it can attract other bodies to itself. We call this attractive force gravity.

 

[CabinetMaker]

So — ignoring air resistance — to the old question of which falls faster, a brick or a feather, we should say the brick. Even if the difference is utterly negligible (exponentially more so if dropped together).

Right?

Right.

 

[gcthomas]

So — ignoring air resistance — to the old question of which falls faster, a brick or a feather, we should say the brick. Even if the difference is utterly negligible (exponentially more so if dropped together).

Right?

Nope. If they were dropped together, then the Earth would move the same for each. Dropped together = same time to fall.

 

[The Barbarian]

Some are offended by the questions. I teach kids as young as 11 sometimes, so I don't mind explaining. Most kids of that age do understand, but not all of them.

1. If the earth's gravity is stronger than that of the moon, how does the moon's gravity over power that of the earth regarding the tides of the ocean?

It doesn't. But the force of the Moon still applies. Hence, the Moon revolves around the Earth, but more precisely both the Earth and the Moon revolve around a point determined by the gravity of both bodies, which is very close to the center of the Earth.

2. Why is it not thought that the sun's gravity causes the tides?

It does. However, because of the inverse square law, the force of gravity from the Sun is less at the surface of the Earth than is the force of the Moon.
The Sun is the cause of neap and spring tides. Neap tides occur when the Moon and Sun are at right angles to each other with respect to the Earth, and Spring tides occur when the Earth, Moon, and Sun are very close to lined up in a row.

3. Why aren't other objects affected by the moon's gravity?

They are. The crust of the Earth, for example, has a measurable strain, due to the Moon's gravity. Not being liquid, it doesn't move very far, though.

4. If the moon's gravity is uniform, why are the tides not uniform?

See above on spring and neap tides.

5. Why aren't ALL bodies of water affected in this way, such as lakes, ponds and puddles?

They are. But the difference in gravity from the Moon on one side of a puddle to the other is so tiny as to be unmeasurable.

True tides—changes in water level caused by the gravitational forces of the sun and moon—do occur in a semi-diurnal (twice daily) pattern on the Great Lakes. Studies indicate that the Great Lakes spring tide, the largest tides caused by the combined forces of the sun and moon, is less than five centimeters in height. These minor variations are masked by the greater fluctuations in lake levels produced by wind and barometric pressure changes.

Consequently, the Great Lakes are considered to be non-tidal.
http://oceanservice.noaa.gov/facts/gltides.html

How can the moon's gravity function in this manner?

Pretty much the way gravity always works. And now you know.

 

[CabinetMaker]

Nope. If they were dropped together, then the Earth would move the same for each. Dropped together = same time to fall.

That is a very interesting problem to solve.

 

[Mocking You]

Let me give it a go.

Every object attracts every other objects to itself. The more mass the object has the more it can attract other bodies to itself. We call this attractive force gravity.

Fixed your explanation.

 

[Stripe]

Nope. If they were dropped together, then the Earth would move the same for each. Dropped together = same time to fall.

To my way of thinking, the Earth would be attracted slightly more toward the object of greater mass, shortening the time for it to impact. However, the difference would be exponentially divided compared with dropping them separately.

Right?

 

[gcthomas]

To my way of thinking, the Earth would be attracted slightly more toward the object of greater mass, shortening the time for it to impact. However, the difference would be exponentially divided compared with dropping them separately.

Right?

That's true if you dropped the objects sequentially, but not if you dropped them at the same time.

 

[Stripe]

That's true if you dropped the objects sequentially, but not if you dropped them at the same time.

Might I respectfully disagree.

To make my thinking clearer, drop the brick and the feather on opposite sides of the Earth.

Repeat the experiment with one dropped on Manhattan and one dropped on LA.

The closer together the places they are dropped from, the more equal their times to impact will be (though, to be fair, the differences would be numbers too stupidly small to even worry about. )

Right?

 

[Stripe]

This discussion is going a whole lot better than it did last time. :chuckle:

:think:

I should hijack threads more often.

 

[gcthomas]

Might I respectfully disagree.

To make my thinking clearer, drop the brick and the feather on opposite sides of the Earth.

Repeat the experiment with one dropped on Manhattan and one dropped on LA.

The closer together the places they are dropped from, the more equal their times to impact will be.

Right?

... Until it is identical when dropped side by side. OK. Is this going anywhere?