NASA launches a satellite into orbit at a height above the surface of the Earth equal to the...

Question:

NASA launches a satellite into orbit at a height above the surface of the Earth equal to the Earth's mean radius. The mass of the satellite is 670 kg. (Assume the Earth's mass is 5.97 1024 kg and its radius is 6.38 106 m.) (a) How long, in hours, does it take the satellite to go around the Earth once? (b) What is the orbital speed, in m/s, of the satellite? (c) How much gravitational force, in N, does the satellite experience?

Gravitation:

Gravitation is the result of the mass of a body. Any massive body attracts another massive body by the force of gravitation. The magnitude of this force depends on the mass of the bodies and the distance between their center of masses. The force of gravitation is responsible for the motion of ball thrown in the air, the motion of planets around the star, motion of satellites around the planets, etc.

Answer and Explanation:

(a) The time period for a satellite moving around a planet is given by Kepler's law of planetary motion. These laws are derived from Newton's law of gravitation and were given by Johannes Kepler in the 17th century. According to the third law of Kepler, the time period (T) of a satellite is given as,

{eq}T = 2\pi \sqrt{\frac{R^3}{GM}} {/eq}

where,

  • R is the distance between the center of masses of satellite and planet
  • G is gravitational constant
  • M is the mass of planet

using the values given in problem and using the value of G as {eq}6.67 \times 10^{-11}\ N.m^2/kg^2 {/eq}

The distance between the center of masses of satellite and planet is given by adding the radius of the planet and the height at which satellite revolves.

Since the satellite is at a height above the surface of the Earth equal to the Earth's mean radius, the distance R is twice the Earth's radius

{eq}T = 2\pi \sqrt{\frac{(2 \times 6.38 \times 10^6)^3}{(6.67 \times 10^{-11})(5.97 \times 10^{24})}}\ s\\ = 14360.2\ s {/eq}

We may convert our answer in hours by realizing that one hour has 3600 s,

time taken by satellite in hours,

{eq}T = 14360.2/3600\ hr\\ = 4.0\ hr {/eq}

(b) The orbital speed (v) of the satellite is given by equating the gravitational force ({eq}F_g {/eq}) and the centripetal force ({eq}F_c {/eq}) on the satellite,

{eq}F_g = F_c\\ \frac{GmM}{R^2} = \frac{mv^2}{R} {/eq}

m is the mass of satellite

solving for v to get,

{eq}v = \sqrt{\frac{GM}{R}} \\ =\sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{2 \times (6.38 \times 10^6)}} \\ = 5586.3\ m/s {/eq}

(c) The gravitational force ({eq}F_g {/eq}) experienced by satellite is given as,

{eq}F_g = \frac{GmM}{R^2}\\ = \frac{(6.67 \times 10^{-11})(670)(5.97 \times 10^{24})}{(2 \times 6.38 \times 10^6)^2}\\ = 1639\ N {/eq}


Learn more about this topic:

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Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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