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لڏپلاڻ · ماڊول 4

Orbital speed: the vis-viva equation

One short equation gives the speed of anything in orbit at any point, if you know how far it is and the size of its orbit.

12 منٽ

NASA/JPL/Space Science Institute

هي سبق Local Solar System فاؤنڊيشن ٽيم جي هڪ ميمبر طرفان تيار ڪيو ويو ۽ ان کي ڪنهن قابل جائزو وٺندڙ طرفان بهتر ڪيو وڃي ٿو. توھان ھن موضوع ۾ ماهر آھيو ڇا؟

مدار

پورو صفحو کوليو
دور
هاڻي رفتار
تيز (پرياپسس)
ھلڪو (آپوپس)
ويجھو
سڀ کان پري
Kepler's 3rd law check: T² / a³

اڻڄاتل هر موڙ کي 8 سيڪنڊن ۾ راند ڪندو آھي؛ ڏيکاريل عرصو حقيقي آهي.

ڊيٽابيس (لکڻيءَ جو متبادل)
مدار جي چوڌاري فاصلو ۽ رفتار بارھ برابر وقت جي قدمن تي
وقت جو قدمسج کان فاصلوهاڻي رفتار

How fast is a spacecraft moving? Tell me how far it is from the planet and how big its orbit is, and one short equation answers the question anywhere along its path. Mission planners use it every day.

Earth seen through the seven windows of the International Space Station's Cupola.
Earth through the Cupola of the International Space Station, which circles the planet about every 90 minutes. Credit: NASA

One short equation gives the speed anywhere along an orbit.

Look

Anything in orbit is always trading height for speed, like a skateboarder on a half-pipe.

  • Falling closer to the central body, it speeds up.
  • Climbing away, it slows down.

In the sandbox the yellow arrow shows the speed: long arrow, fast; short arrow, slow. Watch it grow as the body swings in, and shrink as it climbs out.

Earth does the same thing, just gently because its orbit is almost circular: it moves about 30.3 km/s in early January and about 29.3 km/s in early July.

Understand

The vis-viva equation (“living force”, an old word for energy) gives the speed at any point:

v = √( μ · (2/r − 1/a) )

  • v: speed
  • μ: the central body’s gravitational parameter (GM)
  • r: current distance from the centre of the central body
  • a: semi-major axis of the orbit

Check it:

  • Circle (r = a): v = √(μ/a). That is the circular speed from the Newton lesson.
  • At periapsis, r is smallest, so 2/r is largest and v is fastest.
  • Very far away on an escape path (a → ∞): v = √(2μ/r). That is the escape speed, covered in the next lesson.

Worked example: Earth around the Sun, μ = 1.327 × 10¹¹ km³/s², a = 149.6 million km. At aphelion r = 152.1 million km:

v = √(1.327 × 10¹¹ × (2/1.521 × 10⁸ − 1/1.496 × 10⁸)) ≈ 29.3 km/s.

Master

Vis-viva is energy conservation in disguise. The specific orbital energy (energy per kilogram) is

ε = v²/2 − μ/r = −μ / (2a)

Kinetic plus potential energy is constant along the orbit and depends only on a. Solve for v and you get vis-viva. Consequences:

  • The sign of ε classifies the orbit: negative for ellipses, zero for a parabola, positive for a hyperbola.
  • At any given distance, the speed alone determines a. This is why a single engine burn at one point changes the size of the whole orbit, and why the next lessons can talk about delta-v budgets.

Apsis speeds follow from vis-viva with r = a(1 ± e):

v_p = √(μ/a · (1+e)/(1−e)),   v_a = √(μ/a · (1−e)/(1+e))

Their product is v_p · v_a = μ/a = v_circ², so the circular speed is the geometric mean of the two, not the arithmetic mean. That answers the Try it question.

ڪوشش ڪريو

With the Sun as central body, set a = 1 au and e = 0. Note the speed. Now raise e to 0.6 and compare the fastest and slowest speeds with that number. Is the average of the two equal to the circular speed?

جلدي سوال

3 جلدي سوال. جيڪڏهن توھان صحيح آھيو ته ڏسڻ لاءِ جواب چونڊيو.

  1. In the vis-viva equation, what happens to the speed as the body moves closer to the central body?

    1. A It increases
    2. B It decreases
    3. C It stays constant
    4. D It becomes zero
    جواب ڏيکار

    A. It increases

  2. For a circular orbit (r = a), vis-viva gives:

    1. A v = √(2μ/r)
    2. B v = √(μ/r)
    3. C v = μ/r
    4. D v = 0
    جواب ڏيکار

    B. v = √(μ/r)

  3. Earth moves about 29.8 km/s around the Sun on average. Roughly how fast at perihelion?

    1. A 29.3 km/s
    2. B 30.3 km/s
    3. C 35 km/s
    4. D 42 km/s
    جواب ڏيکار

    B. 30.3 km/s v_p = √(μ/a · (1+e)/(1−e)) with e = 0.0167 gives about 30.3 km/s.

لائين ختم ڪريو

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