Question

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?

Answer 1

Hey There!

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Answer:

`\huge\boxed{Option D}`

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DATA:

Radius of Mercury =
`R_m` =
`2.43x10^6m`

Mass of Mercury =
`M_m = 3.2x10^(23)m`

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

Gravitational Constant =
`G = 6.67x10^(-11) (N.m^2)/(kg^2)`

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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,

`F_g =F_C`

Fg is Given by,

`F_g = (GM_MM_S)/(r^2)`

Fc is Given by,

`F_c=(M_SV^2)/(r)`

Where,

G is Gravitational Constant

`M_e` is mass of Planet Mercury

`M_S` 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

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Now,

`(GM_MM_S)/(r^2) =(M_SV^2)/(r)`

`V = \sqrt(GM_M)/(r)`

r can also be written as,

`V = \sqrt(GM_M)/(R_M +d)`

Substitute the variables,

`V = \sqrt{((6.67x10^(-11))x(3.2x10^(23)))/(2695000)`

Simplify the equation,

V = 2814
`(m)/(s)`

Approximately,

V = 2800
`(m)/(s)`

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Best Regards,

'Borz'