at what altitude should a satellite be placed into circular orbit so that its orbital period is...

Question:

At what altitude should a satellite be placed into circular orbit so that its orbital period is 48 hours? The mass of the earth is 5.976x10^24 kg and the radius of the eatrh is 6.378x10^6 m.

Planets:

When a planets move round the earth due to the gravitational attraction of the earth, then the relation between the time period of the orbit and the radius of the orbit that is given by: {eq}\displaystyle T^2 = \frac{4 \pi^2 a^3}{G M} {/eq} where:

  • {eq}T {/eq} is time taken to compute one orbit that is called time period.
  • {eq}a {/eq} is the distance between the planet and earth that is called the radius of the orbit.
  • {eq}M {/eq} is the mass of the earth.
  • {eq}G = 6.67 \times 10^{-11} {/eq} is the universal gravitational constant.

Answer and Explanation:


Given:

  • The time period of the orbit is: {eq}T = 48 \ \rm hours = 48 \times 60 \times 60 \ \rm sec = 172,800 \ \rm sec {/eq} and the mass of the earth is: {eq}M = 5.976 \times 10^{24} \ \rm kg {/eq}.

As we know the radius of the orbit equals the distance between the satellite and the center of the earth, therefore we can write:

$$\begin{align*} \displaystyle a &= R_e + d &\text{(Where } R_e \text{ is the radius of the earth and } d \text{ is the distance between the satellite and earth)}\\ a &= 6.378 \times 10^6 + d \end{align*}' $$


Now we will compute the radius of the orbit.

As we know the relation between the time period and radius of the orbit is:

$$\begin{align*} \displaystyle T^2 &= \frac{4 \pi^2 a^3}{G M} &\text{(Where } T = 172,800 \ \rm sec, G = 6.67 \times 10^{-11}, a = 6.378 \times 10^6 + d \text{ and } M = 5.976 \times 10^{24} \ \rm kg \text{)}\\ 172,800^2 &= \frac{4 \times 3.14^2 \times (6.378 \times 10^6 + d)^3}{6.67 \times 10^{-11} \times 5.976 \times 10^{24}} &\text{(Plugging in all given values)}\\ 172,800^2 &= \frac{39.4384 (6.378 \times 10^6 + d)^3}{39.85992 \times 10^{13}} \\ 39.4384 (6.378 \times 10^6 + d)^3 &= 172,800^2 \times 39.85992 \times 10^{13} \\ (6.378 \times 10^6 + d)^3 &= \frac{172,800^2 \times 39.85992 \times 10^{13}}{39.4384} &\text{(Dividing both sides by 39.4384)}\\ (6.378 \times 10^6 + d)^3 &= \frac{29.85984 \times 10^{8} \times 39.85992 \times 10^{13}}{39.4384} \\ (6.378 \times 10^6 + d)^3 &= 30.17898 \times 10^{21} \\ 6.378 \times 10^6 + d &= \sqrt[3]{30.17898 \times 10^{21}} &\text{(Taking cube root both sides)}\\ 6.378 \times 10^6 + d &= 3.1134 + 10^7 \\ d &= 31.134 \times 10^6 - 6.378 \times 10^6 \\ d \ &\boxed{ = 24.756 \times 10^6 } \ \rm m \end{align*} $$


Learn more about this topic:

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Planets of the Solar System: Orbits & Visibility

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