Physics, 03.12.2020 17:20, kwilly60

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 mass

is 3.2 x 10^23 kg. What is the spee

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answered: schoolgirl61

Hey There!_____________________________________

$\huge\boxed{Option D}$

_____________________________________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 = $G = 6.67x10^{-11} \frac{N.m^2}{kg^2}$

_____________________________________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,

$F_g = \frac{GM_MM_S}{r^2}$

Fc is Given by,

$F_c=\frac{M_SV^2}{r}$

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

_____________________________________

Now,

$\frac{GM_MM_S}{r^2} =\frac{M_SV^2}{r}$

$V = \sqrt\frac{GM_M}{r}$

r can also be written as,

$V = \sqrt\frac{GM_M}{R_M +d}$

Substitute the variables,

$V = \sqrt{\frac{(6.67x10^{-11})x(3.2x10^{23})}{2695000}$

Simplify the equation,

V = 2814 $\frac{m}{s}$

Approximately,

V = 2800 $\frac{m}{s}$

_____________________________________Best Regards,'Borz'

answered: Guest

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb

answered: Guest

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