# Is it possible that the gravity that keeps our feet planted on the Earth is..

You can't really argue with a crank. They know they're right because they thought of it. You can't show them math, because they don't understand math. None of them have ever taken a physics course, and they're proud of their ignorance. And they all have boobtubes.

This is just another crank.

Show me the math AlexG. Show me the math where you have calculated trajectories of projectiles from Earth without using gravity, but using momentum and inertia instead. Believe me, I am definitely open to being proven wrong. This is driving me crazy.

Show me the math AlexG. Show me the math where you have calculated trajectories of projectiles from Earth without using gravity, but using momentum and inertia instead. Believe me, I am definitely open to being proven wrong. This is driving me crazy.

When you were shown math, your reply was I don't know any math. Again, you say, 'Prove me wrong', the cranks favorite mantra.

It isn't just the rotation of a moon or planet that matters. There are many factors you must take into account. Circumference, orbit speed, rotation speed, orbit circumference, inclination, axial tilt. The amount mass is important when calculating how much surface gravity it has.
Won't fly, no matter how much you throw in. In fact that will just randomize the results. You keep saying "calculate", but you've already said you don't do math, which obliterates the idea that you can say it can be calculated.

The moon doesn't rotate very quickly, but it does revolve around the Earth much more quickly than the Earth revolves around the sun. The inclination of its orbit enables gravity to be on its poles.
I have no idea what that means, but you're not even close. Refer to the gravity map of the moon.

Generally speaking, the faster a planet rotates and the larger its radius, the more gravity it will have and therefore the more massive it can become.
If that were even remotely true, the above map would be flat, and Neil Armstrong would have weighed . . . whatever you throw in there. But he didn't. He weighed just what Newton predicted he would weigh.

That being said. I admit that I don't know how to do the calculations.
Can you multiply? We can get to the bottom of this real quick if you'll pull up a calculator and punch in a few numbers.

When you were shown math, your reply was I don't know any math. Again, you say, 'Prove me wrong', the cranks favorite mantra.

If you don't want to do the math, will you show me how to do the math? The math you showed me was using the gravitational constant. That has to be omitted.

Won't fly, no matter how much you throw in. In fact that will just randomize the results. You keep saying "calculate", but you've already said you don't do math, which obliterates the idea that you can say it can be calculated.

I have no idea what that means, but you're not even close. Refer to the gravity map of the moon.

If that were even remotely true, the above map would be flat, and Neil Armstrong would have weighed . . . whatever you throw in there. But he didn't. He weighed just what Newton predicted he would weigh.

Can you multiply? We can get to the bottom of this real quick if you'll pull up a calculator and punch in a few numbers.

If you go here http://www.1728.org/vectors.htm and take 225km/s at 118 degrees, 29.7 at 0 degrees, and 0.465 at 336.56 degrees, then calculate. you get 212.35km/s at 110.83 degrees.

if you go here http://www.smartconversion.com/unit_calculation/Acceleration_calculator.aspx and calculate 0 to 212.35km/s over 359.0170848 minutes (1/4 Earth day), You get 9.85793383241985m/s^2 Since the 225km/s around the galaxy is approximate and the 29.7km/s around the sun is a mean orbital speed, there is room for adjustment to make that number closer to 9.81m/s^2.

Edit:Actually, If you change the 118 degrees (62 degree incline of ecliptic to solar plane) to 120 degrees (60 degree incline) then you get 211.38 for the resultant and 9.81290347773444m/s^2 for the acceleration.

If you don't want to do the math, will you show me how to do the math?
You're asking how to do math that doesn't work. There is no way, because it doesn't work.
You were shown the math, but you don't want to use the gravitational constant, so you don't want to consider gravity. So take momentum in kg*m/sec and inertia in kg/m[sup]2[/sup] and come up with force in kg*m/sec[sup]2[/sup]

If you go here http://www.1728.org/vectors.htm and take 225km/s at 118 degrees

Why 118 degrees?

, 29.7 at 0 degrees,

Why 0 degrees?

and 0.465 at 336.56 degrees,

Why 336.56 degrees.

then calculate. you get 212.35km/s at 110.83 degrees.

What does the 110.83 degrees represent?

calculate 0 to 212.35km/s over 359.0170848 minutes (1/4 Earth day), You get 9.85793383241985m/s^2

Why did you divide the speed by 1/4 earth day?

Why do you neglect the speed of the Milky Way moving towards Andromeda?

Why do you neglect the speed of the Milky Way moving towards the Virgo Super Cluster?

Why am I wasting my time????

If you go here http://www.1728.org/vectors.htm and take 225km/s at 118 degrees, 29.7 at 0 degrees, and 0.465 at 336.56 degrees, then calculate. you get 212.35km/s at 110.83 degrees.

if you go here http://www.smartconversion.com/unit_calculation/Acceleration_calculator.aspx and calculate 0 to 212.35km/s over 359.0170848 minutes (1/4 Earth day), You get 9.85793383241985m/s^2 Since the 225km/s around the galaxy is approximate and the 29.7km/s around the sun is a mean orbital speed, there is room for adjustment to make that number closer to 9.81m/s^2.

Whoa dude. Hold the phone. Where did the numbers come from and how did you end up with units of acceleration?

Edit:Actually, If you change the 118 degrees (62 degree incline of ecliptic to solar plane) to 120 degrees (60 degree incline) then you get 211.38 for the resultant and 9.81290347773444m/s^2 for the acceleration.

What the -- ?

A perfect example of crank numerology.

Why 118 degrees?

Why 0 degrees?

Why 336.56 degrees.

What does the 110.83 degrees represent?

Why did you divide the speed by 1/4 earth day?

Why do you neglect the speed of the Milky Way moving towards Andromeda?

Why do you neglect the speed of the Milky Way moving towards the Virgo Super Cluster?

Why am I wasting my time????

1/4 Earth day is the maximum amount of time that we would be able to measure the acceleration of something heading towards 1 point on the Earth while it is not going around a curve. I thought this was a logical time to use.

What do you mean why? Those are the speeds the Earth moves and the angles at which it moves at those speeds. I neglected the speeds that the galaxy moves towards Andromeda and the Virgo Super Cluster because the galaxy, and therefore the Solar system, doesn't change directions relative to those often enough to matter. The speed the earth revolves around the galaxy is relative to the speed the galaxy moves towards Andromeda and Virgo.

A perfect example of crank numerology.

But how is it so coincidental? Every single one of those numbers comes from a speed from the Earth.

This is indeed a waste of time. Arguing with a crank always is.

This is indeed a waste of time. Arguing with a crank always is.

WTF man, I give you math and you decide that it's not valid? Why are you even in this thread?

Show how you get the inverse square function. Show how you get gravitational time dilation. Show how you solve the two body problem. Show how you calculate the force.

You take a bunch of selected numbers and massage it until you get a familiar number and say, there's the math.

You solve nothing.

Show how you get force from momentum and inertia, and show that it reflects actual measurement. If you can't do that, you haven't shown anything.

jiveabillion,

I thought you just had some small, albeit fatal, holes in how you think gravity works. But now you've made it impossible for anyone (or anything) to weigh the same amount depending on whether they are standing or reclining (horizontal vs vertical). You've got to be kidding. Plus our weight now changes with time of day and day of year. Plus, you have to contend with the NASA gravity maps. Notice the density changes at the craters affect the measured field strength. Same for the maps of Earth.

For someone who doesn't do math, you whipped out that polar calculator rather handily.

Do you know what units are? You mentioned you're a programmer. Try coding a function that operates on one type and returns another for which there is no conversion, and without throwing an exception.

Worst of all: you don't want to end up with 9.81 m/s² since that will vaporize your little scheme. It was arrived at by applying the law you are trying to defeat.

Show how you get the inverse square function. Show how you get gravitational time dilation. Show how you solve the two body problem. Show how you calculate the force.

You take a bunch of selected numbers and massage it until you get a familiar number and say, there's the math.

You solve nothing.

Show how you get force from momentum and inertia, and show that it reflects actual measurement. If you can't do that, you haven't shown anything.

One thing's for sure. I wouldn't want either you or Origin writing any of my exam questions. You'd actually find out how little I knew.

1/4 Earth day is the maximum amount of time that we would be able to measure the acceleration of something heading towards 1 point on the Earth while it is not going around a curve. I thought this was a logical time to use.

What do you mean why? Those are the speeds the Earth moves and the angles at which it moves at those speeds. I neglected the speeds that the galaxy moves towards Andromeda and the Virgo Super Cluster because the galaxy, and therefore the Solar system, doesn't change directions relative to those often enough to matter. The speed the earth revolves around the galaxy is relative to the speed the galaxy moves towards Andromeda and Virgo.
You realize I'm just humoring you.

If it were even remotely true that the orientation to the center of the galaxy has anything to do with local effects, then we could use an accelerometer like a compass, just turn it toward the center of the galaxy and get a maximum reading.

This is why I never bought into horoscopes when I was young an impressionable. I didn't care so much about classifying people according to which of 12 annual phases they were born; I was just absolutely certain that the mere position of the objects in the sky had ZERO effect on anything on Earth. I confess that had I looked through a telescope quite a bit before I ever even heard of astrology, but observation is like that. It can be quite convincing.

You realize I'm just humoring you.

If it were even remotely true that the orientation to the center of the galaxy has anything to do with local effects, then we could use an accelerometer like a compass, just turn it toward the center of the galaxy and get a maximum reading.

This is why I never bought into horoscopes when I was young an impressionable. I didn't care so much about classifying people according to which of 12 annual phases they were born; I was just absolutely certain that the mere position of the objects in the sky had ZERO effect on anything on Earth. I confess that had I looked through a telescope quite a bit before I ever even heard of astrology, but observation is like that. It can be quite convincing.

I would be more inclined to believe that I was completely wrong if it wasn't for the fact that the planets move so conveniently well to create gravity on all of their surfaces.

Edit: I never said it mattered where the center of the galaxy is. That's just a point of reference. It's all about relative motion. It's more about the motion of everything else relative to whatever object you are tracking.

Nobody has really answered the real question here.

If I can throw a ball backwards from my car and it still follows me, why can't I do that off the back of an Earth with no gravity?

Show me the math AlexG. Show me the math where you have calculated trajectories of projectiles from Earth without using gravity, but using momentum and inertia instead. Believe me, I am definitely open to being proven wrong. This is driving me crazy.
Ok, let's try this. The Earth is accelerating in a direction $$\hat{a}$$. Calculating $$\hat{a}$$ could be hard, depending on your choice of reference frame, but I hope we can agree that there is some direction in which the Earth is accelerating. Because the Earth is a sphere, if you want to jump in a particular direction, you can always do so by choosing the right point on the Earth's surface to stand on when you jump. If you choose the point where your jump would be in the opposite direction of the Earth's acceleration, $$-\hat{a}$$, the Earth would accelerate away from you as you jumped. That would cause you to fly away from the Earth. It doesn't matter how many different kinds of motion you include (orbital, rotational, etc). When you put them all together, the Earth can only be accelerating in one direction at once, so it will always be possible to jump away from that acceleration.

Edit to answer a second question:
Nobody has really answered the real question here.

If I can throw a ball backwards from my car and it still follows me, why can't I do that off the back of an Earth with no gravity?

If you throw a ball backwards from your car, it doesn't "follow" you. It moves away from you at a steady rate, it keeps going straight if you go around a corner, and it certainly doesn't return to your hand after a certain amount of time. By contrast, we return to the Earth if we try to jump off of it, and we follow the Earth's orbital motion. The situations aren't at all similar.

If I can throw a ball backwards from my car and it still follows me,

Ninth grade physics says it doesn't, the distance between the ball and your car increases. Did you really finish high school?

why can't I do that off the back of an Earth with no gravity?

Because you failed basic science.