Initially, the velocity vector is . At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by
, so the velocity is
.
Converting back to direction and magnitude, we get
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:
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