A satellite of mass 210 kg is launched from a site on Earth's equator into an orbit at 210 km...

Question:

A satellite of mass 210 kg is launched from a site on Earth's equator into an orbit at 210 km above the surface of Earth.

(a) Assuming a circular orbit, what is the orbital period of this satellite?

(b) What is the satellite's speed in its orbit?

(c) What is the minimum energy necessary to place the satellite in orbit, assuming no air friction? J

A practice problem on the Earth satellite in circular orbit:

An orbit is a repeated closed path followed by a satellite. A satellite is an object that revolves around another normally larger object in space. There are natural satellites like the moon to the Earth and artificial satellites are manmade objects put into orbits. The artificial satellites are sent to space to orbit any celestial body for various purposes. Based on the intended purpose, the earth satellites are made to orbit at different altitudes and orientations. The orbital velocity of a satellite is inversely proportional to the square root of the orbital radius and the time period of revolution is proportional to the orbital radius raised to the power of three halves. Both these two consequences are due to the inverse square dependence of gravitational force on distance. To launch a satellite from the surface of the Earth to an intended orbit, the required energy needs to be provided to the satellite. The energy need of a satellite includes the energy required to overcome the gravitational potential difference from the surface of the Earth to the orbit and the equivalent kinetic energy to impart orbital velocity necessary to counter the gravitational attraction to stay in orbit. The present problem is concerned mainly with evaluating the energy required to deploy a satellite in a designated orbit.

Answer and Explanation:

An Earth satellite with a mass m (=210 kg) launched from a site on the Earth's equator to orbit the Earth at an altitude {eq}h\ (=2.1 \times 10^5\ m) {/eq} is considered in the problem. The orbit is assumed to be circular. The queries asked are to evaluate the orbital time period, orbital velocity, and minimum energy required to deploy the satellite in the designated orbit. The method followed to find the answers is to equate the centrifugal force of the satellite in the circular orbit to the gravitational force at the orbital distance r ( =R+h = 6378 km + 210 km= 6588 km). This condition is used to valuate the time required to complete a revolution-the orbital period T and the orbital speed {eq}\Large v_{orb} {/eq} as well. The minimum quantity of energy {eq}E_{min} {/eq} required by the satellite to reach from the surface of the Earth to the orbit consists of two components. One is the satellite's potential energy difference on account of its repositioning from the surface of the Earth to the orbit. The second is the kinetic energy as owing to the orbital velocity. Orbital velocity is necessary to counter the attractive gravity force to remain in orbit. The minimum energy requirement is equal to the sum of the potential energy difference and kinetic energy. The sequential steps to reach each solution are itemized below at (a), (b), and (c).

{eq}\text { (a) The orbital period of the satellite : } {/eq}

For the satellite to remain in orbit, the centrifugal force due to the circular motion of the satellite should match the force of gravity at an orbital distance r measured from the center of the Earth.

The force of gravity at orbital radius r:

{eq}\displaystyle F_g=\frac{GMm}{ r^2} \qquad \cdots(1) {/eq}

The centrifugal force of the satellite orbiting at r with angular velocity of revolution {eq}\large \omega_{rev} : {/eq}

{eq}\displaystyle F_c=mr\omega_{rev}^2 = mr \left (\frac{ 2\pi}{T}\right)^2 \qquad \cdots(2) \qquad \text{ ( T is the orbital period ) }\\ Since\ F_c=F_g\\ \displaystyle \frac{mr (2\pi)^2}{T^2}=\frac{GMm}{ r^2}\\ \displaystyle \Rightarrow T= \frac{2\pi}{\sqrt{GM}} r^{\frac{3}{2}} \\ =5.321752 \times 10^3\ sec.{/eq}

{eq}\textbf{The orbital period } \large T=5321.752227\ sec. {/eq}

{eq}\text { (b) The orbital speed of the satellite in circular orbit: } {/eq}

As the satellite is in a circular orbit, the centrifugal force and the force due to gravity at orbital distance r must be equal.

The force of gravity at orbital radius r as at equation - 1:

{eq}\displaystyle F_g=\frac{GMm}{ r^2} {/eq}

The centrifugal force of the satellite in circular orbit at r with orbital velocity {eq}\large v_{orb} {/eq}

{eq}\displaystyle F_c=m\frac{v_{orb}^2}{r} \qquad \cdots(3)\\ \text{Equating } F_c\ and\ F_g \text{ given by equations-1 and 3}\\ \displaystyle m\frac{v_{orb}^2}{r}=\frac{GMm}{ r^2}\\ \text{Rearranging to get } v_{orb}\\ \displaystyle \Rightarrow v_{orb}= \sqrt{ \frac{GM}{r}} \qquad \cdots(4) \\ \hspace 27 pt=7.778194 \times 10^3\ m/sec. {/eq}

{eq}\textbf{The orbital velocity } \large v_{orb}=7.778194\ km/sec. {/eq}

{eq}\text { (c) The minimum energy required to launch the satellite from the surface of the Earth to the orbit } : {/eq}

The energy required to reposition the satellite from the surface of the Earth is equal to the potential energy difference of the satellite at surface and orbit.

{eq}\textbf{The potential energy difference:} {/eq}

The gravitational potential at the surface of the Earth:

{eq}PE_{es} \displaystyle =- \frac{GMm}{R} \qquad \cdots(5) \qquad ( G \text{ is the universal gravitation constant, } M \text{ is the Earth's mass, m is the satellite's mass, and R is the radius of the Earth } ) {/eq}

The gravitational potential energy at the orbit:

{eq}PE_{orb} \displaystyle =- \frac{GMm}{r} \qquad \cdots(6) \qquad (\text{ r is the orbital radius })\\ \text{The difference in gravitational potential energy from surface of the Eath to the orbit:}\\ \Delta PE= PE_{orb}-PE_{es} \displaystyle = -\frac{GMm}{r} +\frac{GMm}{R} \\ \hspace 21 pt \displaystyle =\frac{GMm}{R}-\frac{GMm}{r} \\ \Delta PE \displaystyle =GMm \left \{ \frac{1}{R}-\frac{1}{r} \right \} \qquad \cdots(7) {/eq}

{eq}\textbf {The kinetic energy of the orbiting satellite : } {/eq}

The orbital velocity derived at equation -4 based on the equality of centrifugal and gravity forces at orbital radius r is repeated below.

{eq}\displaystyle V_{orb}=\sqrt{ \frac{GM}{r}}\\ \text{The kinetic energy of the orbiting satellite is:}\\ \displaystyle KE_{orb}=\frac{1}{2}mV_{orb}^2 \qquad ( V_{orb}\ \text{ is the velocity of the satellite in the orbit} )\\ \text{Substituting the expression for } v_{orb}\\ \displaystyle \hspace 26 pt =\frac{1}{2}m \left( \sqrt{ \frac{GM}{r}} \right)^2\\ \displaystyle \hspace 26 pt =\frac{1}{2} \frac{GMm}{r} \qquad \cdots(8)\\ \text{The minimum energy requirement is equal to the increase of the gravitational potential energy plus the orbital kinetic enery.}\\ \displaystyle E_{min}=\Delta PE + KE_{orb} \qquad ( \text { Substituting the expressions for } \Delta PE\ and\ KE_{orb} \text{ from equations- 7 & 8 ) }\\ \hspace 25 pt= GMm \left \{ \frac{1}{R}-\frac{1}{r} \right \} \large+\frac{1}{2} \frac{GMm}{r} \\ \hspace 25 pt = GMm \left \{ \large \frac{1}{R}-\frac{1}{2r} \right\} \qquad \cdots(9)\\ \text{Substituting the values for the universal gravitation constant } G= 6.67408 \times 10^{-11}\ m^3kg^{-1}s^{-2},\ \text{mass of the Earth } M= 5.972 \times 10^{24}\ kg, \text{ equatorial radius of the Earth } R=6.378 \times 10^6\ m,\\ \text{and given radius of orbit } r\ = 6.588\ \times 10^6\ m\ \text{in equation-9}\\ \hspace 18 pt \displaystyle = 6.67408 \times 10^{-11} \times 5.972 \times 10^{24} \times 210 \times \left \{ \frac{1}{6.378 \times 10^6}-\frac{1}{2 \times 6.588\ \times 10^6} \right\} \\ \hspace 18 pt= 6.770855 \times 10^9\ J {/eq}

{eq}\textbf{ The minimum energy required to launch the satellite from a site on equator of the earth to a circular orbit with radius r } ( = 6.588 \times 10^6\ m ) \textbf{ measured from }\\ \textbf{center of the Earth, } E_{min}\ =6.770855 \times 10^9\ J {/eq}


Learn more about this topic:

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How Orbits Are Influenced by Gravity & Energy

from Basics of Astronomy

Chapter 25 / Lesson 6
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