- #1

- 799

- 89

## 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

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?

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?