BRAINLIST!!! ANSWER PLEASE •_• a spaceship orbiting around the earth at a distance 5 of times the radius of the earth from its Centre if the acceleration due to gravity on the earth is 9.8 m s^-2 calculate the acceleration due to gravity acting on the spaceship.
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Answer:
Explanation:
Given that,
Acceleration a=2.45m/s
2
Mass of earth M=6×10
24
kg
Radius of earth R=6×10
6
m
Gravitational constant G=6.67×10
−11
Nm
2
/kg
Now centripetal acceleration is
a
c
=
R
v
2
a
c
=
R
(
R
GM
)
2
a
c
=
R
2
GM
a
c
=
(R+h)
2
GM
Now, put the values
2.45=
(6×10
6
+h)
2
6.67×10
−11
×6×10
24
(6×10
6
+h)
2
=
2.45
6.67×10
−11
×6×10
24
(6×10
6
+h)
2
=
2.45
40.02×10
13
(6×10
6
+h)
2
=1.6334×10
14
6×10
6
+h=
1.6334×10
14
h=1.278×10
7
−6×10
6
h=(12.78−6)×10
6
h=6.78×10
6
h=6.8×10
6
m
So, h≅R
Hence, the height of spaceship is R
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