a) The acceleration of gravity is 
b) The critical velocity is 6668 m/s (24,006 km/h)
c) The period of the orbit is 8452 s
d) The satellite completes 10.2 orbits per day
e) The escape velocity of the satellite is 9430 m/s
f) The escape velocity of the rocket is 11,191 m/s
Explanation:
a)
The acceleration of gravity for an object near a planet is given by

where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
h is the height above the surface
In this problem,
(mass of the Earth)
(Earth's radius)
(altitude of the satellite)
Substituting,

b)
The critical velocity for a satellite orbiting around a planet is given by

where we have again:
(mass of the Earth)
(Earth's radius)
(altitude of the satellite)
Substituting,

Converting into km/h,

c)
The period of the orbit is given by the circumference of the orbit divided by the velocity:

where


v = 6668 m/s
Substituting,

d)
One day consists of:

While the period of the orbit is
T = 8452 s
So, the number of orbits completed by the satellite in one day is

e)
The escape velocity for an object in the gravitational field of a planet is given by

where here we have:



Substituting, we find

f)
We can apply again the formula to find the escape velocity for the rocket:

Where this time we have:


, because the rocket is located at the Earth's surface, so its altitude is zero.
And substituting,

Learn more about gravitational force:
brainly.com/question/1724648
brainly.com/question/12785992
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