Given mmars equals 6.4191 multiplied by 1023 kilograms, rmars equals 3.397 multiplied by 106...


Given mmars = 6.4191 x 1023 kg, rmars = 3.397 x 106 m, G = 6.67428 x 10{eq}^{-11} {/eq} {eq}N-m^2/kg^2 {/eq}, satellite weighs 3100 kg and has 1.9 times radius of mars.

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

Given the force of attraction between mars and satellite is 1348.

The Orbital speed of a satellite moving in a circular motion:

A satellite that is orbiting a planet in a circular motion has an orbital speed.

The orbital speed of the satellite is given by;

{eq}v = \sqrt{\dfrac{GM_p}{r}} {/eq}

where, {eq}M_p {/eq} is the mass of the planet, {eq}r {/eq} is the distance of the satellite from the planet and {eq}G = 6.67 \times 10^{-11} \ Nm^2/kg^2 {/eq} is the gravitational constant.

Answer and Explanation:


  • The mass of mars is {eq}m_{mars} = 6.4191\times 10^{23}\ kg {/eq}
  • Radius of the mars {eq}r_{mars} = 3.397\times 10^6\ m {/eq}
  • The gravitational constant is {eq}G = 6.67428\times 10^{-11}\ Nm^2 kg^{-2} {/eq}
  • Mass of the satellite is {eq}m = 3100\ kg {/eq}
  • The distance of the satellite from the Mars is;

{eq}\begin{align} r &= 1.9 \times r_{mars}\\ &= 1.9\times3.397\times 10^6\ m\\ &=6.4543\times 10^6\ m\\ \end{align} {/eq}

The speed of the satellite is

{eq}\begin{align} v &= \sqrt{\dfrac{Gm_{mars} }{r}}\\ &=\sqrt{\dfrac{6.67428\times 10^{-11} \times 6.4191\times 10^{23}}{6.4543\times 10^6}}\\ &=2.576 \times 10^3 \ m/s\\ \end{align} {/eq}

Learn more about this topic:

Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16

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