The radius of the planet Mercury is 2.43 x 10^6m and its mass
is 3.2 x 10^23 kg. What is the speed of a satellite in orbit
265.000 m above the surface?

The Radius Of The Planet Mercury Is 2.43 X 10^6m And Its Massis 3.2 X 10^23 Kg. What Is The Speed Of

Answers

Answer 1
Hey There!_____________________________________Answer:

[tex]\huge\boxed{Option D}[/tex]

_____________________________________DATA:

Radius of Mercury = [tex]R_m[/tex] = [tex]2.43x10^6m[/tex]

Mass of Mercury = [tex]M_m = 3.2x10^{23}m[/tex]

Distance Satellite above the surface of the Mercury = d = 265,000m

Gravitational Constant =  [tex]G = 6.67x10^{-11} \frac{N.m^2}{kg^2}[/tex]

_____________________________________SOLUTION:

Since the Satellite is orbiting around the Planet Mercury, due to the centripetal force, and Centripetal force is the force that acts towards the center of the circle, Whereas The gravitational force also acts towards the center of the circle thus we can say that Centripetal force is equal or same as centripetal force.  So,

               

                                             [tex]F_g =F_C[/tex]

Fg is Given by,

                                           [tex]F_g = \frac{GM_MM_S}{r^2}[/tex]

Fc is Given by,

                                           [tex]F_c=\frac{M_SV^2}{r}[/tex]

Where,

            G is Gravitational Constant

            [tex]M_e[/tex] is mass of Planet Mercury

            [tex]M_S[/tex] is Mass of Satellite

             r(small letter) is the distance between the center of the Planet Mercury and the satellite.

             V is velocity of satellite

_____________________________________

Now,

                                 [tex]\frac{GM_MM_S}{r^2} =\frac{M_SV^2}{r}[/tex]

                         

                                 [tex]V = \sqrt\frac{GM_M}{r}[/tex]

 r can also be written as,

                           

                                  [tex]V = \sqrt\frac{GM_M}{R_M +d}[/tex]

Substitute the variables,

         

                                  [tex]V = \sqrt{\frac{(6.67x10^{-11})x(3.2x10^{23})}{2695000}[/tex]

Simplify the equation,

           

                                  V = 2814 [tex]\frac{m}{s}[/tex]

Approximately,

                                  V = 2800 [tex]\frac{m}{s}[/tex]

_____________________________________Best Regards,'Borz'

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Answers

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Answers

Answer:

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__

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Answers

Answer:

a

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b

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Explanation:

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Answer:

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Answers

Answer:

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From the question we have

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We have the final answer as

150 kg.m/s

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Answers

Answer:

An airplane cruising at a level height and a steady speed.

A car going along a straight level road at steady speed.

A vibrating spring in a sewing machine.

A ship steaming on a straight course at steady speed.

A train going along the tracks at steady speed.

Explanation:

Hope this helps :)

Answer:

Examples of Uniform Motions.

The hr. ...

An airplane cruising at a level height and a steady speed.

A car going along a straight level road at steady speed.

A vibrating spring in a sewing machine.

A ship steaming on a straight course at steady speed.

A train going along the tracks at steady speed

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