There is a moon in circular orbit of radius 7·108 m around Planet X. How fast must the moon be going?

Question

Given:
Planet X's radius = 2·106 m
Planet X's mass = 9·1022 kg

Answer 1

I think this is what you want.

Using:

G - Universal gravitational constant (6∙7 x 10^-11 N m² Kg^-²).

T² = 4π²d^3 / GM
T² = 4π²(7∙108)^3 / (6∙7x10^-11)(9∙1022)
T² = 14177∙57633 / 6∙098474 x10^-10
T = √(2∙324 7744 15 x10^13)
T = 482 1591∙454 sec. (for moon).

Cir. = 2 π r
Cir. = 2 π (7∙108)
Cir. = 44∙660 88116 m.

Vel. = Cir. / T
Vel. = 44∙660 88116 m / 482 1591∙454 sec.
Vel. = 9∙262 x 10^-6 m/s.
Vel. = 9∙262 µm/s

Are you sure the mass of the planets are typed in correctly?

Answer 2

unexpectedly changing the mass of Saturn does no longer impression its orbit in any respect. human beings look to ask those varieties of questions each and all of the time. they're all according to, dare I say, fairly intense lack of information of orbital mechanics. in fact, many of the solutions are additionally based in intense lack of information. the real fact of the difficulty is that no disruptions are going to electrify the Earth except yet another planet comes VERY VERY on the factor individuals. Jupiter could ought to get far closer to us than Mars with a view to have any disruptive bring about any respect on the earth. A planet the size of Mars could ought to get interior of a few hundred thousand miles to have plenty result. ideas you, throw Jupiter right into a chaotic orbit, and over the process thousands and thousands of years, it may slowly result the the remainder of the photograph voltaic device...yet this could require some heavily close interactions between Jupiter and the different planets.

Answer 3

G = 6.67259 x 10-11 N m2/kg2.
Fg = -GmM/r^2
a = -GM/r^2 = -v^2/r
v^2 = GM/r
v^2 = (6.67259*10^-11)(9*10^22)/(7*10^8)
v^2 = 8579 m^2/s^2
v = 92.623 m/sec