Satellite orbit formula, physical satellite formula

Updated on science 2024-03-27
12 answers
  1. Anonymous users2024-02-07

    If the satellite is moving in a uniform circular motion, it can be calculated as "v = root number gm r (r is the distance from a certain point to the earth)".

    The satellite moves in a uniform circular motion because the centripetal force satisfies: f=gmm rr=mvv r.Now it's going to be in elliptical orbit.

    Choose a point as the orbit change point, and accelerate the satellite at this point so that its velocity becomes (v+dv), so that its speed does not meet the formula:

    f=gmm rr=mvv r, the speed is high, it will be centrifuged. So it becomes a circle that is not the original circumference. On the earth, it is raised, and the potential energy increases.

    Then the speed decreases. [The initial orbit change point is called perigee] and when it reaches the apogee point, the speed is not enough to meet the orbital speed of the place [small], so it falls back [near the center of the earth]. Return to perigee.

    And so on and so forth, running on an elliptical orbit.

    Not only at perigee, but also at apogee, the linear velocity is not equal to the local orbital velocity, and the other points are not equal to the other points.

    Calculation method: Calculate with the conservation of mechanical energy. If the position of the change in potential energy is not considered, and the gravitational acceleration changes, it is easy to calculate, and the orbital velocity can be calculated from the intersection point of the minor axis first, and then the other points can be calculated by the conservation of mechanical energy; If you want to take it into account, you need to use the integral calculation.

    When starting a change of orbit, if the speed is reduced, the point is apogee.

  2. Anonymous users2024-02-06

    The relationship between the speed and altitude of a satellite is relatively simple:

    The gravitational potential energy of an object on the earth is e=gmm r, that is to say, the energy of an object on the earth escaping from the earth = gmm r (r is equal to the radius of the earth on the ground, and the distance to the center of the earth is equal to in space).

    Using this formula, it is easy to conclude that the second cosmic velocity is equal to the root number 2 * first cosmic velocity =

    So let's say the launch speed of the satellite is VO

    Then 1 2MVO 2 = GMM R-GMM (R+H)+1 2MV 2 (M is the mass of the Earth, G is the gravitational constant, and H is the height above the ground).

    So v = root number (1 2vo 2-gm r + gm r + h).

    gmm r-gmm (r+h) (in fact, the exact gravitational potential energy that overcomes the gravitational pull of the earth from the ground to h).

    From the formula gmm r 2=mv 2 r to v 2=gm r, it can be seen that the larger the radius, the greater the velocity.

    Then from the formula v=2 to r t, we can see that the smaller the velocity, the larger the period.

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  4. Anonymous users2024-02-04

    Formula: GM R2=(2 T)2R=G=W2R

    The basic formula for the motion of physical celestial bodies:

    1.Kepler's Law of the Third.

    t2 r3=k(=4r2 gm)t: period, k: constant (independent of the mass of the planet, depending on the mass of the central object).

    2.The law of gravitation is in vain.

    f=gm1m2 r2 (g=, in their line).

    3.Gravity and gravitational acceleration on celestial bodies.

    gmm/r2=mg;g=gm r2 (r: celestial radius (m), m: celestial mass (kg)).

    4.Satellite orbiting speed, angular velocity.

    Period: v=(gm r)12; w=(gmr3)12;t = 2r(r3 gm)1 2(m: central celestial mass).

    5.1 (2 & 3) Cosmic velocity v1 = (g r ) 1 2 (gmr ) 1 2 =; v2=;v3=

    6.Geosynchronous satellite GMM (R+H)2=M4R2(R+H) T2(H=36000km).

  5. Anonymous users2024-02-03

    Ideological method: geostationary satellites.

    That is, the cycle of movement is the same as that of the earth. Let's first push to the universal substitution formula, the earth's surface: mg=gmm r 2 gm=gr 2 (universal substitution formula) geostationary satellite:

    m'w^2(r+h)=gmm'(r+h) 2 (r+h) 3=gm w 2=gr 2 w 2 h=(gr w) to the third power root-rThen the pure punch bent w=t 2

  6. Anonymous users2024-02-02

    The orbital radius of a geostationary satellite can be calculated by the following formula:

    r=gm/v^2=

    where r represents the orbital radius of the synchronous satellite, g represents the number of constant macrofronts of the retracement force, m represents the mass of the ground drain carrying ball, and v represents the velocity of the satellite.

  7. Anonymous users2024-02-01

    Satellite orbit parameters, also known as satellite orbit roots, are various parameters used to describe the position, shape, and orientation of a satellite in space.

    Orbital plane inclination: the angle between the equatorial plane and the satellite orbit plane, which is calculated by the counterclockwise rotation from the equatorial nuclear slag plane to the orbital plane when the satellite orbit ascends.

    Altitude: The distance of the satellite from the Earth's surface.

    Substellar point: The intersection point of the satellite and the Earth's center line on the Earth's surface.

    Ascending node: The intersection of the satellite's trajectory at the equator from south to north.

    Period: The time it takes for a satellite to make one orbit around the Earth.

    Intercept: The degree by which the satellite orbits the Earth.

    Eccentricity: Focal length vs. orbital half-length.

  8. Anonymous users2024-01-31

    If the satellite moves in a uniform circular motion around the earth, the satellite operation period can be calculated by gmm (r 2) = 4 ( 2) m (t 2);

    A satellite is a celestial body that orbits a planet, star, or other celestial body. They maintain orbital stability through the interaction between gravity and the celestial bodies around them. Satellites can be natural or artificial.

    Natural satellites are formed by gravitational capture of planets, stars, or other celestial bodies. For example, the natural satellite of the Earth is the Moon, while the natural satellite of Mars is the Phobos.

    First, Phobos II class.

    Artificial satellites are artificial devices that humans make and send into space. There are many types of artificial satellites, including communication satellites, meteorological satellites, navigation satellites, scientific research satellites, etc. Piki Bridges are used for various purposes, such as providing global communication services, monitoring weather changes, navigation and positioning, earth observation, space exploration, etc.

    The operation of satellites depends on Newton's laws of gravity and Kepler's laws of planetary motion. By adjusting the speed and orbit of a satellite, one can keep the satellite in a stable orbit to achieve a variety of different goals and tasks. <>

  9. Anonymous users2024-01-30

    Are you here to deduce it yourself or do you need a conclusion from the direct source.

    If you want to derive :bai

    You need du to have university knowledge)zh

    1. Abandon Kepler's law, because of the undergraduate.

    It has been rigorously proven in post-DAO astrophysics that Kepler's law is an approximate exception and cannot withstand rigorous astrodynamic analysis.

    2. The motion of the two bodies of the celestial body is equivalent to the movement of a single body around a constant central body.

    3. Using the long and short axes as the coordinates, the motion vector projection analysis is carried out, and the differential equations are listed and solved. (It must be able to be solved rigorously).

    If you want an answer directly, keep asking.

  10. Anonymous users2024-01-29

    1.Kepler's third law: t2 r3 k( 4 2 gm) {r: orbital radius, t: period, k: constant (independent of the mass of the planet, depending on the mass of the central object)}

    2.The law of gravitation: f gm1m2 r2 (g, direction on their line).

    3.Gravity and acceleration due to gravity on celestial bodies: gmm R2 mg; g=gm r2 (r: celestial radius (m), m: celestial mass (kg)).

    4.Satellite orbit velocity, angular velocity, period: v (gm r)1 2;ω=gm/r3)1/2;t 2 (r3 gm)1 2{m:central celestial mass}

    5.1 (2 & 3) Cosmic velocity v1 = (g r ) 1 2 (gmr ) 1 2 =; v2=;v3=

    6.Geosynchronous satellites gmm (R+H)2 m4 2(R+h) t2{H 36000km, H: height above the Earth's surface, R: radius of the Earth}

    Note: (1) The centripetal force required for the motion of celestial bodies is provided by gravitational force, f to f thousand;

    2) the law of gravitation can be applied to estimate the mass density of celestial bodies;

    3) Geosynchronous satellites can only operate over the equator, and the period of operation is the same as that of the Earth's rotation;

    4) The orbital radius of the satellite becomes smaller, the potential energy becomes smaller, the kinetic energy becomes larger, the velocity becomes larger, and the period becomes smaller (together with three antis); (5) The maximum orbital velocity and the minimum launch velocity of the earth satellite are both.

  11. Anonymous users2024-01-28

    p-cycle.

    a Semi-major axis.

    The scale factor can be found from the circular motion.

  12. Anonymous users2024-01-27

    Same as the circular motion formula, except that r is changed to a

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