[Yahya (OP)]

I'm not sure about the numbers

Then leave it to NASA.
They are sure, because they did the experiment- repeatedly.

when ? where ? could you please give me a link to that experiment ? anything ?

[PmbPhy]

U= − GMm/r
The first mass m is supposed to be at infinity that means it exerts some potential energy U to escape from the surface of mass M with radius r and approach infinity , when mass m is at infinity ....

That's not possible since infinity is not a place but a concept. See:
https://en.wikipedia.org/wiki/Infinity

Infinity (symbol: ∞) is a concept describing something without any bound or larger than any natural number.
....

[Kryptid]

whether slower or faster, I claim no actual test for escape velocity.

Then why did you claim that rockets are launched far above escape velocity?

Also, you haven't explained why hydroelectric power plants don't produce twice as much power as they are expected to or why elevators don't consume twice as much electricity as they are expected to.

[Bored chemist]

I'm not sure about the numbers

Then leave it to NASA.
They are sure, because they did the experiment- repeatedly.

when ? where ? could you please give me a link to that experiment ? anything ?

Have you forgotten so soon?

They did the test.
Repeatedly.
It was called the Apollo project.
They actually did launch stuff from the moon.
It behaved the way that equations (and everybody but you) say it should.

[Yahya (OP)]

I'm not sure about the numbers

Then leave it to NASA.
They are sure, because they did the experiment- repeatedly.

when ? where ? could you please give me a link to that experiment ? anything ?

Have you forgotten so soon?

They did the test.
Repeatedly.
It was called the Apollo project.
They actually did launch stuff from the moon.
It behaved the way that equations (and everybody but you) say it should.

A link ?

[Colin2B]

They did the test.
Repeatedly.
It was called the Apollo project.
They actually did launch stuff from the moon.
It behaved the way that equations (and everybody but you) say it should.

A link ?

https://www.nasa.gov/audience/forstudents/5-8/features/nasa-knows/what-was-apollo-program-58.html

Lots of info here.

[Yahya (OP)]

They did the test.
Repeatedly.
It was called the Apollo project.
They actually did launch stuff from the moon.
It behaved the way that equations (and everybody but you) say it should.

A link ?

https://www.nasa.gov/audience/forstudents/5-8/features/nasa-knows/what-was-apollo-program-58.html

Lots of info here.

A quote?

[Bored chemist]

How much spoonfeeding  of the obvious do you need before you accept that it's not "the rest of the world" who made a mistake; it's you.

[Yahya (OP)]

How much spoonfeeding  of the obvious do you need before you accept that it's not "the rest of the world" who made a mistake; it's you.

to a level you get the information from a quote in any site.

[Colin2B]

They did the test.
Repeatedly.
It was called the Apollo project.
They actually did launch stuff from the moon.
It behaved the way that equations (and everybody but you) say it should.

A link ?

https://www.nasa.gov/audience/forstudents/5-8/features/nasa-knows/what-was-apollo-program-58.html

Lots of info here.

A quote?

Lazybones, do your own research. I personally don't care if you put out theories that make you look an idiot, not my job to ensure you don't. Your role is to seach for and identify all related data and take it into account.
Listen to @Bored chemist he knows what he's talking about.

[Kryptid]

There is more than one way to calculate gravitational potential energy. Here is another way to do it.

Let’s say you have a 1 kilogram ball and you drop it from a height of 1 meter. What is the final velocity of the ball just before it touches the ground? We can use the following equation to find out:

V_f = -√(2ah), where

“V_f” is the final velocity in meters per second,
“a” is the gravitational acceleration at the Earth’s surface, and
“h” is the height above the Earth’s surface that the ball is dropped from.

So now we’ll put in the relevant numbers and calculate the final velocity:

V_f = -√((2)(9.8 m/s²)(5 m))
V_f = -√(98)
V_f = -9.8995 meters per second

The value is negative because it is a vector quantity (it is moving towards the Earth).

Here is a video detailing how the equation can be derived:

The next question is, “how much kinetic energy did the ball have at its final velocity”? We can use the following kinetic energy equation to calculate this:

E_k = (½)mv², where

“E_k” is the kinetic energy in joules,
“m” is the mass of the ball in kilograms, and
“v”  is the velocity of the ball in meters per second

Now to solve the equation:

E_k = (½)(1 kg)((9.8995 m/s)²)
E_k = (½)(1)(98)
E_k = 49 joules

However, since this is energy being released instead of being absorbed, we would give the value a negative sign. The energy is therefore -49 joules.

The next question becomes, which equation is this result consistent with, the original gravitational potential energy equation or your proposed revision of it? Let’s find out. First, I’ll use the original equation:

U = -G(Mm/r)
U = -(6.674 x 10^-11 m³ kg^-1 s^-2)((5.97237 x 10²⁴ kg)(1 kg)/(6,371,005 meters))
U = -(6.674 x 10^-11 m³ kg^-1 s^-2)(9.3742981 x 10¹⁷)
U = -62,564,065.4496 joules

This calculation is the amount of energy released when the 1 kilogram ball falls from “infinity” down a height 5 meters above the Earth’s surface, so this is not the entire story. Since our ball is falling from 5 meters above the Earth’s surface, we need to find the difference between this value and the value of gravitational potential energy right at Earth’s surface:

U = -G(Mm/r)
U = -(6.674 x 10^-11 m³ kg^-1 s^-2)((5.97237 x 10²⁴ kg)(1 kg)/(6,371,000 meters))
U = -(6.674 x 10^-11 m³ kg^-1 s^-2)(9.3742981 x 10¹⁷)
U = -62,564,114.5503 joules

Now we subtract these two values from each other:

(-62,564,114.5503 joules) - (-62,564,065.4496 joules) = -49.1007 joules

We know that your equation just inserts a “2”, so the value it would calculate would be -98.2014 joules.

So it turns out that the original equation is consistent with the prior calculations. Your equation posits that the correct value should be twice that (about -98 joules), so your equation is not consistent with the gravitational potential calculation.

[Bored chemist]

to a level you get the information from a quote in any site.

Sorry, but that doesn't make sense.