Answer
given,
mass of satellite = 545 Kg
R = 6.4 x 10⁶ m
H = 2 x 6.4 x 10⁶ m
Mass of earth = 5.972 x 10²⁴ Kg
height above earth is equal to earth's mean radius
a) satellite's orbital velocity
centripetal force acting on satellite = 
gravitational force = 
equating both the above equation
v = 5578.5 m/s
b) 
T = 14416.92 s
T = 4 hr
c) gravitational force acting
F = 5202 N
a
a
b
b
a
b
a
This will really help you learn a lot.
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
