A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ______ 10^10 J. (Mass…

Question

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ______

\times 10^{10} J

.
(Mass of earth

= 6 \times 10^{24} kg

, Radius of earth

= 6.4 \times 10^{6} m

, Gravitational constant

= 6.67 \times 10^{- 11} Nm^{2} kg^{- 2}

)

Accepted Answer

Correct answer: 3. (3) The kinetic energy (KE) of a satellite revolving around the Earth can be expressed with the following formula: KE= 12mv^2= 12m GM_er= GM_em2r For this satellite, the radius r is the sum of the Earth's radius R_E and the height h above the Earth's surface: r=R_E+h Given: Mass of the satellite, m=1000 kg Mass of the Earth, M_e=6 10^24 kg Radius of the Earth, R_E=6.4 10^6 m Height of orbit above the Earth's surface, h=270 km=2.7 10^5 m Gravitational constant, G=6.67 10^-11 Nm^2 kg^-2 Substitute these values into the equation: KE= 6.67 10^-11 6 10^24 10002(6.4 10^6+2.7 10^5) = 6.67 10^-11 6 10^24 10002 6.67 10^6 =3 10^10J

Solution

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