satellite revolving around the earth at height of 400 above the earth surface has time ?
Answers
Explanation:
Mass of the Earth,
M=6.0 \times 10^{24}\ kg
M=6.0×10
24
kg
m=200\ kg
m=200 kg
R_e=6.4 \times 10^6\ m
R
e
=6.4×10
6
m
G=6.67 \times 10^{-11}\ Nm^2kg^{-2}
G=6.67×10
−11
Nm
2
kg
−2
Height of the satellite,
h=400\ km = 4 \times 10^5\ m
h=400 km=4×10
5
m
Total energy of the satellite at height h
= (1/2)mv^2+(-GM_em/(R_e+h))
=(1/2)mv
2
+(−GM
e
m/(R
e
+h))
Orbital velocity of the satellite,
v=\sqrt{\dfrac{GM_e}{R_e+h}}
v=
R
e
+h
GM
e
Total energy at height h
=\dfrac{1}{2}\dfrac{GM_em}{R_e+h}-\dfrac{GM_em}{R_e+h}
=
2
1
R
e
+h
GM
e
m
−
R
e
+h
GM
e
m
Total\ Energy=-\dfrac{1}{2}\dfrac{GM_em}{R_e+h}
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)
=\dfrac{GM_em}{2(R_e+h)}
=
2(R
e
+h)
GM
e
m
=\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2(6.4 \times 10^6+4 \times 10^5)}
=
2(6.4×10
6
+4×10
5
)
6.67×10
−11
×6×10
24
×200
=5.9 \times 10^9\ J
=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 \times 10^9J
5.9×10
9
J
of energy to just escape it.
Mass of the Earth,
M=6.0 \times 10^{24}\ kg
M=6.0×10
24kg
m=200 kg
R_e=6.4 \times 10^6\ m
Re=6.4×10
6m
G=6.67 \times 10^{-11}\ Nm^2kg^{-2}
G=6.67×10−11Nm
2kg−2
Height of the satellite,
h=400\ km = 4 \times 10^5\ m
h=400 km=4×10
5m
Total energy of the satellite at height h
= (1/2)mv^2+(-GM_em/(R_e+h))
=(1/2)mv
2+(−GMem/(Re+h))
Orbital velocity of the satellite,
v=\sqrt{\dfrac{GM_e}{R_e+h}}
v=Re+hGMe
Total energy at height h
=\dfrac{1}{2}\dfrac{GM_em}{R_e+h}-\dfrac{GM_em}{R_e+h}
=21Re+hGMem−Re+hGMem
Total\ Energy=-\dfrac{1}{2}\dfrac{GM_em}{R_e+h}
Total Energy=−21Re+hGMem
The negative sign indicates that the satellite is bound to the Earth.
Energy required to send the satellite out of its orbit = – (Bound energy)
=\dfrac{GM_em}{2(R_e+h)}
=2(Re+h)GMem
=\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2(6.4 \times 10^6+4 \times 10^5)}
=2(6.4×106+4×105)6.67×10−11×6×1024×200
=5.9 \times 10^9\ J
=5.9×10^9J
If the satellite just escapes from the gravitational field, then total energy of the satellite is zero. Therefore, we have to supply
5.9 \times 10^9J
5.9×10
9
J
of energy to just escape it.
Read more on Brainly.in - https://brainly.in/question/15132368#readmoreMass of the Earth,
M=6.0 \times 10^{24}\ kg
M=6.0×10
24
kg
m=200\ kg
m=200 kg
R_e=6.4 \times 10^6\ m
R
e
=6.4×10
6
m
G=6.67 \times 10^{-11}\ Nm^2kg^{-2}
G=6.67×10
−11
Nm
2
kg
−2
Height of the satellite,
h=400\ km = 4 \times 10^5\ m
h=400 km=4×10
5
m
Total energy of the satellite at height h
= (1/2)mv^2+(-GM_em/(R_e+h))
=(1/2)mv
2
+(−GM
e
m/(R
e
+h))
Orbital velocity of the satellite,
v=\sqrt{\dfrac{GM_e}{R_e+h}}
v=
R
e
+h
GM
e
Total energy at height h
=\dfrac{1}{2}\dfrac{GM_em}{R_e+h}-\dfrac{GM_em}{R_e+h}
=
2
1
R
e
+h
GM
e
m
−
R
e
+h
GM
e
m
Total\ Energy=-\dfrac{1}{2}\dfrac{GM_em}{R_e+h}
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)
=\dfrac{GM_em}{2(R_e+h)}
=
2(R
e
+h)
GM
e
m
=\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2(6.4 \times 10^6+4 \times 10^5)}
=
2(6.4×10
6
+4×10
5
)
6.67×10
−11
×6×10
24
×200
=5.9 \times 10^9\ J
=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 \times 10^9J
5.9×10
9
J
of energy to just escape it.
Read more on Brainly.in - https://brainly.in/question/15132368#readmore