משתמשת:אמא של גולן/טיוטת נוסחאות

נוסחאות המעבר

בהנתן שני מסלולים מעגליים סביב אותו מוקד משיכה, המעבר במסלול הוהמן הוא המסלול האליפטי המשיק לשני מסלולים אלו. נניח ש $r=r_{1}$ הוא רדיוס המסלול הפנימי ו-$r=r_{2}$ הוא רדיוס המסלול החיצוני

עבור גוף קטן קשיח המקיף גוף אחר For a small body orbiting another, very much larger body (such as a satellite orbiting the earth), the total energy of the body is the sum of its kinetic energy and potential energy, and this total energy also equals half the potential at the average distance $a$, (the semi-major axis):

$E={\begin{matrix}{\frac {1}{2}}\end{matrix}}mv^{2}-{\frac {GMm}{r}}={\frac {-GMm}{2a}}.\,$

Solving this equation for velocity results in the Vis-viva equation,

$v^{2}=\mu \left({\frac {2}{r}}-{\frac {1}{a}}\right)$

where:

• $v\,\!$ is the speed of an orbiting body
• $\mu =GM\,\!$ is the standard gravitational parameter of the primary body, assuming $M+m$ is not significantly bigger than $M$ (which makes $v_{M}\ll v$)
• $r\,\!$ is the distance of the orbiting body from the primary focus
• $a\,\!$ is the semi-major axis of the body's orbit.

Therefore the delta-v required for the Hohmann transfer can be computed as follows, under the assumption of instantaneous impulses:

$\Delta v_{1}={\sqrt {\frac {\mu }{r_{1}}}}\left({\sqrt {\frac {2r_{2}}{r_{1}+r_{2}}}}-1\right)$,

to enter the elliptical orbit at $r=r_{1}$ from the $r_{1}$ circular orbit

$\Delta v_{2}={\sqrt {\frac {\mu }{r_{2}}}}\left(1-{\sqrt {\frac {2r_{1}}{r_{1}+r_{2}}}}\,\,\right)$,

to leave the elliptical orbit at $r=r_{2}$ to the $r_{2}$ circular orbit where $r_{1}$ and $r_{2}$ are, respectively, the radii of the departure and arrival circular orbits; the smaller (greater) of $r_{1}$ and $r_{2}$ corresponds to the periapsis distance (apoapsis distance) of the Hohmann elliptical transfer orbit. The total $\Delta v$ is then:

$\Delta v_{total}=\Delta v_{1}+\Delta v_{2}.$

Whether moving into a higher or lower orbit, by Kepler's third law, the time taken to transfer between the orbits is:

$t_{H}={\begin{matrix}{\frac {1}{2}}\end{matrix}}{\sqrt {\frac {4\pi ^{2}a_{H}^{3}}{\mu }}}=\pi {\sqrt {\frac {(r_{1}+r_{2})^{3}}{8\mu }}}$

(one half of the orbital period for the whole ellipse), where $a_{H}\,\!$ is length of semi-major axis of the Hohmann transfer orbit.

In application to traveling from one celestial body to another it is crucial to start maneuver at the time when the two bodies are properly aligned. Considering the target angular velocity being

$\omega _{2}={\sqrt {\frac {\mu }{r_{2}^{3}}}}$

angular alignment α (in radians) at the time of start between the source object and the target object shall be

$\alpha =\pi -\omega _{2}t_{H}=\pi \left(1-{\frac {1}{2{\sqrt {2}}}}{\sqrt {\left({\frac {r_{1}}{r_{2}}}+1\right)^{3}}}\,\right)$

Example

Total energy balance during a Hohmann transfer between two circular orbits with first radius $r_{p}$ and second radius $r_{a}$

Consider a geostationary transfer orbit, beginning at r₁ = 6,678 km (altitude 300 km) and ending in a geostationary orbit with r₂ = 42,164 km.

In the smaller circular orbit the speed is 7.73 km/s; in the larger one, 3.07 km/s. In the elliptical orbit in between the speed varies from 10.15 km/s at the perigee to 1.61 km/s at the apogee.

The Δv for the two burns are thus 10.15 − 7.73 = 2.42 and 3.07 − 1.61 = 1.46 km/s, together 3.88 km/s.[1]

It is interesting to note that this is greater than the Δv required for an escape orbit: 10.93 − 7.73 = 3.20 km/s. Applying a Δv at the LEO of only 0.78 km/s more (3.20−2.42) would give the rocket the escape speed, which is less than the Δv of 1.46 km/s required to circularize the geosynchronous orbit. This illustrates that at large speeds the same Δv provides more specific orbital energy, and energy increase is maximized if one spends the Δv as soon as possible, rather than spending some, being decelerated by gravity, and then spending some more (of course, the objective of a Hohmann transfer orbit is different).

נוסחאות SRP

a_(〖SRP〗_ECI )=-4.55∙〖10〗^(-6) (C_r A)/M s ̂_ECI מיצוע חישוב הפרעות הSRP (לפי ההרצאות שניתנו) באלמנטי מסלול קלאסים 〈da/dt〉=0

〈de/dt〉=-3 fa/2h sinα_sp

〈dω/dt〉=3 fa/2he cosα_sp כאשר f הוא היטל כח השמש על מישור הלווין. אנו מניחים כי הפאנלים מאונכים לשמש (מכוונים במלואם לשמש) ולכן הכח מחושב במלואו. α_sp= λ_s-ω-Ω