# Scientists want to place a 2600-kg satellite in orbit around Mars. They plan to have the...

## Question:

Scientists want to place a 2600-kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 1.7 times the radius of Mars above the surface of the planet. Here is some information that will help solve this problem:

mass (mars) = {eq}6.4191 \times 10^{23} {/eq} kg, radius (mars) = {eq}3.397 \times 10^6 {/eq} m, G = {eq}6.67428 \times 10^{-11} {/eq} N{eq}\cdot {/eq}m{eq}^2 {/eq}/kg{eq}^2 {/eq}

1) What is the force of attraction between Mars and the satellite?

2) What speed should the satellite have to be in a perfectly circular orbit?

3) How much time does it take the satellite to complete one revolution?

4) Which quantities would change the speed the satellite needs to orbit at?

5) What should the radius of the orbit be (measured from the center of Mars), if we want the satellite to take 8 times longer to complete one full revolution of its orbit?

## Orbital Speed:

Orbital speed is the speed by which an object or body in space sweeps or covers the distance of its orbit in a particular time period. Orbital speed depends on the mass of the body and its radius. Its value is expressed in meter per second.

Given data

• The mass of satellite is: {eq}M = 2600\;{\rm{kg}}{/eq}
• The mass of mars is: {eq}{m_m} = 6.4191 \times {10^{23}}\;{\rm{kg}}{/eq}
• The radius of mars is: {eq}{r_m} = 3.97 \times {10^6}\;{\rm{m}}{/eq}
• The gravitational constant is: {eq}G = 6.674 \times {10^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}}{/eq}

(1)

The expression to calculate the force of attraction between mars and satellite,

{eq}F = \dfrac{{GM{m_m}}}{{{r^2}}}{/eq}

Here, {eq}r{/eq} is the distance between satellite and earth.

Calculate the value of {eq}r{/eq} .

{eq}\begin{align*} r &= 1.7 \times {r_m}\\ r &= 1.7 \times 3.97 \times {10^6}\\ r &= 6.749 \times {10^6}\;{\rm{m}} \end{align*}{/eq}

Substitute the value in the above expression,

{eq}\begin{align*} F &= \dfrac{{6.674 \times {{10}^{ - 11}} \times 2600 \times 6.4191 \times {{10}^{23}}}}{{{{\left( {6.794 \times {{10}^6}} \right)}^2}}}\\ F& = 2413.14\;{\rm{N}} \end{align*}{/eq}

Thus, the force of attraction between the mars and satellite is {eq}2413.14\;{\rm{N}}{/eq}

(2)

The expression to calculate the orbit speed is given as,

{eq}V = \sqrt {\dfrac{{G{m_m}}}{r}} {/eq}

Substituting the value in the above expression,

{eq}\begin{align*} V &= \sqrt {\dfrac{{6.674 \times {{10}^{ - 11}} \times 6.4191 \times {{10}^{23}}}}{{6.749 \times {{10}^6}}}} \\ V& = \sqrt {6.6 \times {{10}^6}} \\ V &= 2569.04\;{\rm{m/s}} \end{align*}{/eq}

Thus, the orbital speed is {eq}2569.04\;{\rm{m/s}}{/eq}

(3)

The expression to calculate the time taken to complete one revolution,

{eq}T = \dfrac{d}{V}{/eq}

Here, {eq}d{/eq} is the distance.

Calculate the value of distance.

{eq}\begin{align*} d& = 2\pi r\\ d& = 6.28 \times 6.749 \times {10^6}\\ d& = 42.38 \times {10^6}\;{\rm{m}} \end{align*}{/eq}

Substitute the value in the above expression,

{eq}\begin{align*} T &= \dfrac{{42.38 \times {{10}^6}}}{{2569.04}}\\ T &= 16498.075\;{\rm{s}} \end{align*} {/eq}

Thus, the total time taken is {eq}16498.075\;{\rm{s}} {/eq}

(4)

The quantity would change the speed of satellite is the distance between the satellite and the mars, as the distance increase, orbital velocity will decrease.

(5)

According to the Kepler law,

{eq}{r^3} \propto {T^2} {/eq}

Now, the time period is 8 time longer than,

{eq}\begin{align*} {r^3}& \propto {\left( {8T} \right)^2}\\ {r^3}& \propto 64{T^2}\\ r& \propto 4\sqrt T \end{align*}{/eq}

Hence the radius would be four time greater than the pervious one. Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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