Question

To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were observed by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit--a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass M.For all parts of this problem, where appropriate, use G for the universal gravitational constant.Part A)Find the orbital speed v for a satellite in a circular orbit of radius R. (Express the orbital speed in terms of G, M, and R),Part B)Find the kinetric energy K of a satellite with mass m in a circular orbit with radius R. (Express your answer in terms of m, M, G, and R).Part C)Find the orbital period T. (Express your answer in terms of G, M, R, and pi).Part D)Find L, the magnitude of the angular momentum of the satellite with respect to the center of the planet. (Express your answer in terms of m, M, G, and R.Part E)The quantities v, K, U, and L all represent physical quantities characterizeing the orbit that depend on radius R. Indicate the exponent (power) of the radial dependence of the absolute value of each. (Express your answer as a comma-separated list of exponents corresponding to v, K, U, and L, in that order.

Answer 1

Answer:

a)  v = √ G M / r, , b) K = ½ m G M / r, , c)  T = 2 π √ (r³ / GM)

, d)  L = m √ (rGM)

Explanation:

The formula of universal gravitation

     F = G m M / r²

Part A) orbital speed

Let's use Newton's second law

    F = m a

Where the acceleration is centripetal

    a = v² / r

Let's replace

    G m M / r² = m v² / r

    v = √ G M / r

Part B)

Kinetic energy

     K = ½ m v²

     K = ½ m G M / r

Part C)

The orbital period The speed (speed module) is constant, so we can use the relationships

    v = d / t

The distance in the length of the circle

    d = 2π r

    T = t = 2π r / (√ G M / r) = 2π r √(r / GM)

    T = 2 π √ (r³ / GM)

Part D) The angular momentum

     L = I w

The satellite is small, so we can approximate it to a particle

     I = m r²

The angular and linear velocity are related

     v = w r

    w = v / r

    w = √(G M / r) (1 / r)

We replace

    L = m r² 1 / r √(GM / r)

    L = m r √ (GM / r)

    L = m √ (rGM)

Part E) let's calculate the order of magnitude of the quantities

    v = √G M / r

The value of the distance from the plant center to the satellite

    r = Re + h

the height of the satellites is less than 1 10⁶m from the surface of the earth,

    r = 6.37 10⁶ + 1 10⁶ = 7.37 10⁶

    v = √ (6.67 10⁻¹¹ 5.98 10²⁴ / 7.37 10⁶)

    v = √ (5.4 10⁶)

    v ~ 10³ m / s

    K = ½ m G M / r

The satellite mass about 10⁴ kg

    K = ½ 10⁴ 6.67 10⁻¹¹ 5.98 10²⁴ / 7.37 10⁶

    K ~ 10¹¹ J

    U = G m M / r

    U ~ 10¹¹ J

    L = m √ (rGM)

    L = 10⁴ √ (7.37 10⁶ 6.67 10⁻¹¹ 5.98 10²⁴)

    L ~ 10¹⁴ kg m2 / s