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Navigator · Qaybta 9

Hohmann transfers: changing orbits

To go higher you speed up, and then you arrive slower. The two-burn manoeuvre that moves satellites from low orbit to geostationary orbit.

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NASA/JPL/Space Science Institute

Waxbarashadan waxaa qoray Local Solar System Foundation kooxda xubna ah oo ay ku hagaajin karaan qiimeeye xirfad leh. Ma waxaad u qalmin in ay arrimo this? Soo jeedin edit

Here is the strangest rule of orbital flight: to catch up you slow down, and to go higher you speed up, after which you end up going slower. Walter Hohmann, a German city engineer, worked out in 1925 the cheapest way to move between two circular orbits. Almost every satellite in geostationary orbit got there using his idea.

From a 300 km orbit to geostationary orbit, 35,786 km up: two burns, 5.3 hours. Computed with the vis-viva equation.

Speed up to climb, and you arrive slower than you left.

Look

  1. Burn 1, forwards. The extra speed does not make the satellite go faster around its circle: it stretches the circle into an ellipse. The low point stays where the burn happened; the far side rises to the target height.
  2. Coast. Half an orbit with no engine. The satellite climbs and, like a ball thrown up, slows down as it goes.
  3. Burn 2, forwards again. At the top it is now too slow to stay there and would fall back. A second push makes the orbit circular at the new height.

From 300 km to geostationary orbit the two burns add up to about 3.9 km/s, and the trip takes about 5 hours 17 minutes.

Understand

The speed at any point of an orbit comes from vis-viva: v = √(μ(2/r − 1/a)). A Hohmann transfer ellipse touches both circles, so its semi-major axis is a_t = (r₁ + r₂)/2.

Radius (km)Speed (km/s)
Circular at 300 km6,6787.73
Transfer, low point6,67810.15
Transfer, high point42,1641.61
Circular, geostationary42,1643.07

Burn 1 = 10.15 − 7.73 ≈ 2.43 km/s. Burn 2 = 3.07 − 1.61 ≈ 1.47 km/s. The coast is half the transfer ellipse’s period: π√(a_t³/μ) ≈ 5.3 hours.

(Real geostationary launches also have to change the orbit’s tilt from the launch site’s latitude to the equator, which costs more; launching near the equator saves fuel.)

Master

Δv₁ = √(μ/r₁)·(√(2r₂/(r₁+r₂)) − 1),   Δv₂ = √(μ/r₂)·(1 − √(2r₁/(r₁+r₂)))

The Hohmann transfer is the minimum two-impulse transfer between coplanar circular orbits, as long as r₂/r₁ < 11.94. Beyond that ratio a bi-elliptic transfer (three burns, going beyond the target first) costs less Δv, at the price of much more time.

Burning at the low point is also the most efficient place to add energy: the Oberth effect says a given Δv changes the orbital energy most where the spacecraft is moving fastest, because Δε ≈ v·Δv.

Ku day

Use vis-viva to check the speed at the top of the transfer: μ = 398,600 km³/s², r = 42,164 km, a = 24,421 km. Then compare it with the circular speed at 42,164 km. The difference is the second burn.

Imtixaan degdeg ah

3 su'aalo degdeg ah. Dooro jawaab si aad u aragto haddii aad xaq tahay.

  1. To reach a higher circular orbit, the first burn is:

    1. A Straight up
    2. B Forwards, speeding up
    3. C Backwards, slowing down
    4. D Sideways, out of the orbit plane
    Tuso jawaabta

    B. Forwards, speeding up

  2. During the coast between the two burns, the satellite:

    1. A Speeds up as it climbs
    2. B Slows down as it climbs
    3. C Keeps the same speed
    4. D Stops at the top
    Tuso jawaabta

    B. Slows down as it climbs Climbing away from Earth trades kinetic energy for potential energy, like a ball thrown upwards.

  3. Why is the geostationary orbit useful?

    1. A It is closest to Earth
    2. B A satellite there takes one day per orbit, so it stays over the same spot
    3. C There is no gravity there
    4. D It is the fastest orbit
    Tuso jawaabta

    B. A satellite there takes one day per orbit, so it stays over the same spot

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