Physics question on revolutions, radial acceleration?

Question

An earth satellite moves in a circular orbit with an orbital speed of 5950 m/s. Find the time of one revolution.
Take the gravitational constant to be G = 6.67×10^−11 N*m^2/kg^2, the mass of the Earth to be m_e = 5.97×1024 kg

Find the radial acceleration of the satellite in its orbit.

How would you proceed? details thanks

I followed those procedures but I still came up with the wrong answer....some calculation error i keep making

Answer 1

(G* m1*m2)/r.^2=force on the satellite due to earth in the radial direction ==m1* accelaration of the satellite in radial direction

m1=mass if the satellite
m2 is the mass of earth

accelaration of satellite= v.^2/r

===>G*m2/r=v.^2

==>r=G*m2/v.^2
once r is calculated here by substituting the values og G ,m2,& v ==5950 m/s

time for one revolution == (2* pi *r)/v

Answer 2

v=5950 m/s
G = 6.67×10^−11 N*m^2/kg^2
m_e = 5.97×1024 kg
m = the mass of the satellite (unknown in this case)

From F= ma, the gravitational force = G*m_e*m / r^2
and the a = v^2/r
we can cancel the m from both side of the equation. This lead to:
G*m_e / r^2 = v^2 / r

Therefore,
the radius of the satelite, r = G*m_e / v^2
the time for one revolution = 2*pi*r / v

radial acceleration = v^2/r

Answer 3

Find Radial Acceleration

Answer 4

To find the time it takes to complete one orbit, you just divide distance by velocity:

(2*pi*r)/v = t

That means you need to know the radius of the satellite. If you're in a circular orbit, your NET radial acceleration has to be zero (if the radius stays the same all the way around the circle, then there's no radial acceleration, is there?). That means your centripetal acceleration has to be equal to the acceleration due to gravity:

(G * m_e)/r^2 = v^2/r

which can be simplified to: (G*m_e)/v^2 = r

Now you have the radius and plug it into your first equation to find the time.

The 'official' way to find the period (the way that works for elliptical orbits, as well) is to use Kepler's third law:

t^2 = (4 * pi^2 * a^3)/(G*m_e)

where 'a' is your semi-major axis. For a circular orbit, the radius and semi-major axis are the same. If you rearranged your second equation to solve for v and then substituted it into your first equation, you'd see Kepler's third law and the first equation are equivalent to each other (for circular orbits, at least).