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 =6.0×1024 kg; radius of the earth =6.4×106m; G = 6.67×10−11 Nm2kg−2.

Question

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 =6.0×1024 kg; radius of the earth =6.4×106m; G = 6.67×10−11 Nm2kg−2.

Filo · Accepted Answer

Mass of the Earth, M=6.0×1024 kgm=200 kg Re​=6.4×106 mG=6.67×10−11 Nm2kg−2Height of the satellite,h=400 km=4×105 mTotal energy of the satellite at height h=(1/2)mv2+(−G(Re​+h)Me​m​)Orbital velocity of the satellite, v=Re​+hGMe​​​Total energy at height h =21​Re​+hGMe​m​−Re​+hGMe​m​Total Energy=−21​Re​+hGMe​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(Re​+h)GMe​m​=2(6.4×106+4×105)6.67×10−11×6×1024×200​=5.9×109 JIf the satellite just escapes from the gravitational field, then total energy of the satellite is zero. Therefore, we have to supply 5.9×109J of energy to just escape it.

Question Text	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 $= 200 k g$ ; mass of the earth $= 6.0 \times 1 0^{24} k g$ ; radius of the earth $= 6.4 \times 1 0^{6} m$ ; G = $6.67 \times 1 0^{- 11} N m^{2} k g^{- 2}$ .
Answer Type	Text solution:1
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