Iranian Space program

[rmi5]

You need to wonder and think a little more. It is good for your brain. The negative sign in the equation that I had mentioned earlier shows that for a satellite potential energy is always larger than kinetic energy

View attachment 190865

The negative value is not an algebraic sign. In gravitational equations, Potential energy is shown as negative and kinematic energy as positive as the sum for any given object should be constant (U+K=C). Kinematic energy is always positive so potential energy must be negative for the sum to remain constant. One transfers to the other.

Example :

In case our baseline is earth's surface, a 1kg mass, 1m above earth surface potential energy is -mgh=-9.81 j

Now the same mass at 2m above surface has a potential energy of -19.62 j. Hopefully you are not claiming that the one at 1m height has a higher potential energy. You need to compare the absolute values. ABS(19.62)>ABS(9.81)

What you are talking about?
Potential energy itself, can be summed with any constant.
I don't remember your formula, but if it was assumed that the potential energy is zero at infinity(meaning that your constant is zero), then you would always derive potential energy as a negative number! You can sum your potential energy with a constant=+1e999999 as well. Then you would always derive potential energy as a positive number.
To break it down for you:
F=-grad(U) >> U can be derived as:
-integral of F.dl + U0
for two masses: F=-Gm1m2/r^2, hence
U=-Gm1m2/r+U0
Now, if as r goes to infinity, we assume U=0, then U0=0. But, it is not the only solution. You can also assume that as r=radius of Earth, U=0, then U0=Gm1m2/(radius of the earth)

what matters is:
delta E = delta K + delta U
assuming no friction, then delta E=0

Now if r1>r2 then U(r1) >U(r2). and U(ground)= a constant less than both U(r1), and U(r2).
so, deltaU(from ground to r1) > deltaU(from ground to r2)
Consequently, you would need to have more kinetic energy change if you want to send it to r1 compared to r2.

Example :

In case our baseline is earth's surface, a 1kg mass, 1m above earth surface potential energy is -mgh=-9.81 j

Now the same mass at 2m above surface has a potential energy of -19.62 j. Hopefully you are not claiming that the one at 1m height has a higher potential energy. You need to compare the absolute values. ABS(19.62)>ABS(9.81)

You are very lucky that you are not in my vicinity. Other wise, you would have received some physical punishment
if you assume that U(ground)=0, then delta U(from ground to h) = U(h) = +mgh
U(2)>U(1)>0

 

[Arminkh]

What you are talking about?
Potential energy itself, can be summed with any constant.
I don't remember your formula, but if it was assumed that the potential energy is zero at infinity(meaning that your constant is zero), then you would always derive potential energy as a negative number! You can sum your potential energy with a constant=+1e999999 as well. Then you would always derive potential energy as a positive number.
To break it down for you:
F=-grad(U) >> U can be derived as:
-integral of F.dl + U0
for two masses: F=-Gm1m2/r^2, hence
U=-Gm1m2/r+U0
Now, if as r goes to infinity, we assume U=0, then U0=0. But, it is not the only solution. You can also assume that as r=radius of Earth, U=0, then U0=Gm1m2/(radius of the earth)

what matters is:
delta E = delta K + delta U
assuming no friction, then delta E=0

Now if r1>r2 then U(r1) >U(r2). and U(ground)= a constant less than both U(r1), and U(r2).
so, deltaU(from ground to r1) > deltaU(from ground to r2)
Consequently, you would need to have more kinetic energy change if you want to send it to r1 compared to r2.



You are very lucky that you are not in my vicinity. Other wise, you would have received some physical punishment
if you assume that U(ground)=0, then delta U(from ground to h) = U(h) = +mgh
U(2)>U(1)>0

You contradicted yourself up there:

Yes, in case of a satellite the potential energy decreases as r increases so a satellite at a higher orbit has a lower potential energy compared to a satellite at a lower orbit. The reason is the reduction in gravity. So despite what you said in your fist quote the potential energy doesn't increase as r increases. On earth proximity the gravity is considered constant and that's why potential energy increases with the height.

On the other hand the kinetic energy also decreases when r increases as the speed of the satellite is proportionate to 1/r^0.5.

So overall, your total mechanical energy decreases as you go to a higher orbit which is exactly what this formula shows:

My question was, then why do we need stronger rocket (more energy) to put a payload into higher orbit?

First understand the question then if you know the answer, talk.

 

[rmi5]

You contradicted yourself up there:

Yes, in case of a satellite the potential energy decreases as r increases so a satellite at a higher orbit has a lower potential energy compared to a satellite at a lower orbit.

Nooooo!!! How did you conclude that?
satellite at a higher orbit has higher Potential energy

The reason is the reduction in gravity.

???? Nooo

On earth proximity the gravity is considered constant and that's why potential energy increases with the height.

Do you wanna kill me bro?!!!

On the other hand the kinetic energy decreases when r increases as the speed of the satellite is proportionate to 1/r^0.5.

So overall, your total mechanical energy decreases as you go to a higher orbit which is exactly what this formula shows:

View attachment 190875

My question was, then why do we need stronger rocket (more energy) to put a payload into higher orbit?

First understand the question then if you know the answer, talk.

You are really a typical Bachcheh porrou. I am trying to help you here, but, you are still doing porrou baazi.

 

[SOHEIL]

There's nothing wrong with copying/borrowing , even the american space X is a copy of russian Soyuz

 

[Arminkh]

Nooooo!!! How did you conclude that?
satellite at a higher orbit has higher Potential energy

???? Nooo

Do you wanna kill me bro?!!!


You are really a typical Bachcheh porrou. I am trying to help you here, but, you are still doing porrou baazi.

Do you read what you write? Read what you have written again.

What you are talking about?
Potential energy itself, can be summed with any constant.
I don't remember your formula, but if it was assumed that the potential energy is zero at infinity(meaning that your constant is zero), then you would always derive potential energy as a negative number!

We are not talking about any potential energy. Gravitational potential energy is negative because it is negative work i.e. the force is applied in the opposite direction of the displacement. It has nothing to do with what the potential energy at infinity is.

To break it down for you:
F=-grad(U) >> U can be derived as:
-integral of F.dl + U0
for two masses: F=-Gm1m2/r^2, hence
U=-Gm1m2/r+U0
Now, if as r goes to infinity, we assume U=0, then U0=0. But, it is not the only solution. You can also assume that as r=radius of Earth, U=0, then U0=Gm1m2/(radius of the earth)

Or you can simply substitute g in formula U=mgh by GM/r^2 and h with r and drive at the same equation for potential energy as everything is being measured from the center of the earth. It will still be negative as it is negative work. And we have nothing to do with infinity.

what matters is:
delta E = delta K + delta U
assuming no friction, then delta E=0

Now if r1>r2 then U(r1) >U(r2). and U(ground)= a constant less than both U(r1), and U(r2).
so, deltaU(from ground to r1) > deltaU(from ground to r2)
Consequently, you would need to have more kinetic energy change if you want to send it to r1 compared to r2.

You are very lucky that you are not in my vicinity. Other wise, you would have received some physical punishment
if you assume that U(ground)=0, then delta U(from ground to h) = U(h) = +mgh
U(2)>U(1)>0

No, gravitational potential energy is always negative thus U(from ground to h)=U(h)-U0= -mgh - 0 = -mgh

For two different heights where H1<H2, Delta U (H1 to H2) = -mgH2 - (-mgH1)

The result is a negative value which is correct meaning going from H1 to H2 will increase your potential energy

However, Delta U(H2 to H1) will yield a positive number meaning you will loose potential energy going from H2 to H1 or gain kinetic energy.

Going back to satellite, the potential energy formula is -GMm/r.

Where r2>r1 Delta U (r1 to r2) = -GMm/r2 - (-GMm/r1) = GMm (-1/r2 + 1/r1) which will be positive as r2>r1 meaning we have actually lost potential energy.

At extreme, when r approaches infinity, your potential energy approaches zero while still negative not because we have assumed it is zero at infinity but because r is in denominator.

 

[The SC]

You should emphasize the difference between the launch vehicle and the satellite. Calculations are different, and you should make it clear which one you do not understand. From your original question, I understood that you want to know about the increased energy demand for the launch vehicle, so stick with it and do not get carried away by a random answer concerning something else such as the satellite itself.

 

[rmi5]

Do you read what you write? Read what you have written again.




We are not talking about any potential energy. Gravitational potential energy is negative because it is negative work i.e. the force is applied in the opposite direction of the displacement. It has nothing to do with what the potential energy at infinity is.



Or you can simply substitute g in formula U=mgh by GM/r^2 and h with r and drive at the same equation for potential energy as everything is being measured from the center of the earth. It will still be negative as it is negative work. And we have nothing to do with infinity.



No, gravitational potential energy is always negative thus U(from ground to h)=U(h)-U0= -mgh - 0 = -mgh

For two different heights where H1<H2, Delta U (H1 to H2) = -mgH2 - (-mgH1)

The result is a negative value which is correct meaning going from H1 to H2 will increase your potential energy

However, Delta U(H2 to H1) will yield a positive number meaning you will loose potential energy going from H2 to H1 or gain kinetic energy.

Going back to satellite, the potential energy formula is -GMm/r.

Where r2>r1 Delta U (r1 to r2) = -GMm/r2 - (-GMm/r1) = GMm (-1/r2 + 1/r1) which will be positive as r2>r1 meaning we have actually lost potential energy.

At extreme, when r approaches infinity, your potential energy approaches zero while still negative not because we have assumed it is zero at infinity but because r is in denominator.

No offense, but you have an extremely low IQ. Again you are wrong from the 1st post to the last one. I don't like to repeat myself, so read my previous post again.
Although I doubt if it helps you, since you don't understand basic concepts about negative numbers or about the concept of potential.
My last hint for you, If you put a particle in higher potential, in absence of everything else then it would try to go to lower potential. It's the most basic concept that does not need much brain to understand. so, if U(ground)=0, then U(h) cannot be -mgh, since -mgh<0=U(0) meaning that you would have been thrown from ground to higher heights by the gravity, while the opposite is true, and objects fall from higher heights toward ground by gravity.

 

[mister]

don't squabble please
If we consider the center of gravity, center of the earth, then we calculate the mechanical energy of the Earth's surface; (1) since with increasing distance from the surface of the Earth, gravitational potential energy is reduced and linear velocity to stay in orbit decreases, thus the kinetic energy is reduced, therefore mechanical energy is lower than ground surface. (2) With greater the distance from the ground, mechanical energy is again reduced. We see that the difference between mechanical energy of case 1 and the ground less than the difference between mechanical energy of case 2 and ground. Work (energy) required to put the satellite in an orbit is equal to negative net mechanical energy of the satellite in the orbit (difference between mechanical energy of the orbit and the ground). Therefore it is evident that more energy is required to inject the satellite in higher orbits .

 

[India defense]

Iran is true Islamic nations leader......congratulations

 

[Al-zakir]

Iran launches home-made Safir Fajr GPS satellite into space

Mubarak my brothers. Great news. Allahu Akbar.

 

[500]

2 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 231.2 km
Apogee: 476.7 km
Inclination: 55.5 °
Period: 91.5 minutes
Semi major axis: 6724 km

5 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 229.6 km
Apogee: 463.2 km
Inclination: 55.5 °
Period: 91.3 minutes
Semi major axis: 6717 km

6 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 228.8 km
Apogee: 447.3 km
Inclination: 55.5 °
Period: 91.1 minutes
Semi major axis: 6709 km

7 Feb:

NORAD ID: 40387
Int'l Code: 2015-006A
Perigee: 227.9 km
Apogee: 440.4 km
Inclination: 55.5 °
Period: 91.1 minutes
Semi major axis: 6705 km

 

[Arminkh]

You should emphasize the difference between the launch vehicle and the satellite. Calculations are different, and you should make it clear which one you do not understand. From your original question, I understood that you want to know about the increased energy demand for the launch vehicle, so stick with it and do not get carried away by a random answer concerning something else such as the satellite itself.

Thanks, that may be what I'm missing.

 

[Arminkh]

don't squabble please
If we consider the center of gravity, center of the earth, then we calculate the mechanical energy of the Earth's surface; (1) since with increasing distance from the surface of the Earth, gravitational potential energy is reduced and linear velocity to stay in orbit decreases, thus the kinetic energy is reduced, therefore mechanical energy is lower than ground surface. (2) With greater the distance from the ground, mechanical energy is again reduced. We see that the difference between mechanical energy of case 1 and the ground less than the difference between mechanical energy of case 2 and ground. Work (energy) required to put the satellite in an orbit is equal to negative net mechanical energy of the satellite in the orbit (difference between mechanical energy of the orbit and the ground). Therefore it is evident that more energy is required to inject the satellite in higher orbits .

I agree with all you mentioned above.

So are you implying that we have maximum potential energy on earth compared to center of the gravity and thus the most negative mechanical energy and what the rocket does is to reduce this mechanical energy when it puts the satellite in the orbit?

That makes sense to me. My mistake was that I considered the mechanical energy on surface zero and was wondering why the potential energy is decreasing the higher we go. Thanks!

 

[Arminkh]

No offense, but you have an extremely low IQ. Again you are wrong from the 1st post to the last one. I don't like to repeat myself, so read my previous post again.
Although I doubt if it helps you, since you don't understand basic concepts about negative numbers or about the concept of potential.
My last hint for you, If you put a particle in higher potential, in absence of everything else then it would try to go to lower potential. It's the most basic concept that does not need much brain to understand. so, if U(ground)=0, then U(h) cannot be -mgh, since -mgh<0=U(0) meaning that you would have been thrown from ground to higher heights by the gravity, while the opposite is true, and objects fall from higher heights toward ground by gravity.

The correct answer is up there if you want to learn something. Gravitational potential energy is negative, is maximum on the surface and does decrease the higher you go.

I appreciate your good intentions trying to answer my question. Not knowing something is not a bad thing. However you lack the ability to recognize when you don't know something. We Iranians call this compound ignorance. I have also noticed your tendency to attribute your own defects to others. For example you call others fools while you are the only fool there. This is called physiological projection. There is link below for you to know more about it. You may also want to visit a psychologist and you may as well thank me if and when you are cured.

Psychological projection - Wikipedia, the free encyclopedia

 

[yavar]

Iran Safir-e-Fajr satellite launch images of 4 camera on board /Fajr satellite separation in space