2. At an orbital height of 400 km, find the
orbital period of the satellite.
Answers
Explanation:
PHYSICS
A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth's gravitational influence? Mass of the satellite =200 kg; mass of the earth =6.0×10
24
kg; radius of the earth =6.4×10
6
m; G = 6.67×10
−11
Nm
2
kg
−2
.
November 22, 2019avatar
Mudra Spoorthi
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VIDEO EXPLANATION
ANSWER
Mass of the Earth, M=6.0×10
24
kg
m=200 kg
R
e
=6.4×10
6
m
G=6.67×10
−11
Nm
2
kg
−2
Height of the satellite,h=400 km=4×10
5
m
Total energy of the satellite at height h=(1/2)mv
2
+(−G
(R
e
+h)
M
e
m
)
Orbital velocity of the satellite, v=
R
e
+h
GM
e
Total energy at height h =
2
1
R
e
+h
GM
e
m
−
R
e
+h
GM
e
m
Total Energy=−
2
1
R
e
+h
GM
e
m
The negative sign indicates that the satellite is bound to the Earth.
Energy required to send the satellite out of its orbit = – (Bound energy)
=
2(R
e
+h)
GM
e
m
=
2(6.4×10
6
+4×10
5
)
6.67×10
−11
×6×10
24
×200
=5.9×10
9
J
If the satellite just escapes from the gravitational field, then total energy of the satellite is zero. Therefore, we have to supply 5.9×10
9
J of energy to just escape it.
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Answer:
93 minutes
Explanation:
height of satellite (h) = 400 km
= 400000 m
Radius of the earth (R) = 6370 km
= 6370000 m.
Gravitational constant (a) = 6.673 x 10⁻¹¹ Nm² kg⁻²
Mass of the earth (M) = 5.972 x 10²⁴ kg
v = √Gm/(R + h)
v = √6.673 x 10⁻¹¹ x 5.972 x 10²⁴/6370000 + 400000
v = √39.83 x 10¹³/6770 x 10³
= √5.8 x 10⁷
= 10³ √58
= 7620 m/s.
T = 2π (R + h)/V
= 2 x 22/7 x 6370000 + 400000/7620
= 44/7 x 6770000/7620
= 297880000/53340
= 55845 (60 seconds = 1 minutes)
= 93 minutes