# Why launching closer to the equator constitutes an advantage

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## Summary:

I would like to discuss why launching closer to the equator is physically the most efficient approach.
I recently realized that some of the most used Space Centers (for instance the Kennedy Space Center, Merritt Island, FL (US), by NASA, the Guiana Space Center ,French Guiana, by ESA and the Baikonur Space Center ,Kazakhstan's area leased to Russia, by Roscosmos) are relatively close to the Equator and I was wondering why this is advantageous.

I first thought that it may be related to energy efficiency and I indeed found a similar analysis here.

$$\frac{\Delta E}{m} = \frac{E_f - E_i}{m} = \frac{E_f - (\mathrm{KE}_i + \mathrm{PE}_i)}{m}$$
$$= -\frac{GM}{r_f} - \frac{1}{2} v_i^2 + \frac{GM}{r_i} \\ = \frac{GM}{R_E} \left( 1 - \frac{1}{\alpha} \right) - \frac{1}{2} \left( \frac{2 \pi R \sin \theta}{T} \right)^2 \tag{1}$$

Where

$$PE=-\int_{\infty}^{r} \vec F \cdot d \vec r'=-\int_{\infty}^{r} - \frac{GMm}{r'^2} \hat r' \cdot d \vec r' = -\frac{GMm}{r}$$

$$r_i=R_E, \ r_f= \alpha R_E, \ \alpha \ \text{is finite and greater than 1}$$

$$v_i=\frac{\text{arc length}}{time}=\frac{2 \pi R \sin \theta}{T}$$

Where I've used the standard definition of latitude i.e.

And I assumed ##\theta## is small (as we are close to the equator), so that the chord is (approximately) equal to the arc length.

So based on the above analysis we see that the required energy per unit mass is maximum at the equator, which a priori means that launching from the equator is inefficient in terms of energy! (for those who check the link: tparker uses the an unconventional definition of latitude; he measures it from one of the poles; that is why he gets minimum energy at the Equator).

OK so at this point I thought 'well so there must be another factor that compensates the inefficiency in terms of energy'. I suspect that is the fact that objects on the equator (i.e. at ##\theta=0##) get the fastest rotational speed of any other ##\theta##, which means they need to use less fuel. Does that really compensate?

etotheipi

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$$v_i=\frac{\text{arc length}}{time}=\frac{2 \pi R \sin \theta}{T}$$
If vi is the initial speed than it is

$$v_i=\frac{\text{arc length}}{time}=\frac{2 \pi R \cos \theta}{T}$$

with the definition of latitude you are using.

I suspect that is the fact that objects on the equator (i.e. at ##\theta=0##) get the fastest rotational speed of any other ##\theta##, which means they need to use less fuel.
That't what the amended equation above says. That means it is already included into the calculation. Other factors are the reduced gravity at the equator (due to the flattening of Earth) and the fact that many satellites go into equatorial orbits.

etotheipi and JD_PM
kuruman
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I always thought it's because the apparent weight of the rocket is less nearer the equator because of the centrifugal force that is zero at the poles and increases as you go south or north fro there.

@DrStupid thanks, now everything makes sense! (i.e. the required energy per unit mass is minimum at the equator).

I always thought it's because the apparent weight of the rocket is less nearer the equator because of the centrifugal force that is zero at the poles and increases as you go south or north fro there.
Mmm I've read a bit on this.

Gravity would be weaker on the equator due to the centrifugal force i.e.

$$g_{\text{eff}}:= g - \omega^2 R_E$$

Plugging numbers into we get ##g_{\text{eff}} \sim 3 \times 10^{-2} m/s^2##. However the experimental result is ##g_{\text{eff}} \sim 5 \times 10^{-2} m/s^2##. Such a difference is justified by means of the centrifugal force.

jbriggs444
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I always thought it's because the apparent weight of the rocket is less nearer the equator because of the centrifugal force that is zero at the poles and increases as you go south or north fro there.
I do not think that it is helpful to adopt the rotating frame and consider potential resulting from the centrifugal force field. That potential diverges as one gets far from the Earth, so any conservation of energy argument is wasted.

I do not think that it is helpful to adopt the rotating frame and consider potential resulting from the centrifugal force field. That potential diverges as one gets far from the Earth, so any conservation of energy argument is wasted.
Of course the total energy is not conserved because the rotating frame is not inertial. However, the centrifugal force is conservative. Therefore it shouldn't be a problem to use the corresponding potential ##- {\textstyle{1 \over 2}}\omega ^2 r^2##.

jbriggs444
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Of course the total energy is not conserved because the rotating frame is not inertial. However, the centrifugal force is conservative. Therefore it shouldn't be a problem to use the corresponding potential ##- {\textstyle{1 \over 2}}\omega ^2 r^2##.
Energy is conserved in the rotating frame. However, potential energy is not zero at infinity. It is, instead, infinite.

etotheipi
Energy is conserved in the rotating frame.
If you include the potential of the centrifugal force, than yes. I was under the impression that you don't want to do that.

However, potential energy is not zero at infinity. It is, instead, infinite.
I don't see the problem. The kinetic energy of a body at rest in an inertial system goes positive infinite in the rotating frame and the potential energy goes negative infinite. That fits together. Do I miss something?

etotheipi
kuruman
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I asked about this a friend who is at JPL and he said that launch sites vary depending on the orbit, trajectory and mission. Near-equatorial sites are used for geosynchronous orbits because they provide a relatively higher tangential velocity component while for polar orbits launch sites are farther up north to minimize the tangential component, which is not needed, thus saving fuel.

vanhees71, Dale and etotheipi
jbriggs444
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If you include the potential of the centrifugal force, than yes. I was under the impression that you don't want to do that.
Doing it is fine. Expecting it to be useful in a computation of energy needed to escape is questionable. If you need infinite energy to escape to infinity and you have infinite energy coming from the centrifugal field, how is a little bit of a boost to start with supposed to help?

etotheipi
To achieve orbit, you need to achieve orbital velocity (in inertial space). Whatever it is. Say 18000 mph (in inertial space). If you launch from the pole, the Earth rotation rate contributes zero to your velocity (in inertial space) If you launch with the Earth rotation, you are already going over 1000 MPH in inertial space, (to see this take the circumference of the Earth approx 25000 mile and divide by 24 hours; hence > 1000 mph)

You now need 18000 mph - 1000 mph or 17000 mph. Note if you launch the wrong way, you now need to add 19000 mph to get to 18000 mph (inertial).

Expecting it to be useful in a computation of energy needed to escape is questionable. If you need infinite energy to escape to infinity and you have infinite energy coming from the centrifugal field, how is a little bit of a boost to start with supposed to help?
Rockets are usually not lauched to infinity. Most payloads go into Earth orbits. If a rotating frame is useful or not depends on the circumstances. In case of a geosynchronous orbit it would be my 1st choice.

jbriggs444