A satellite of mass 280 kg is transferred from circular orbit radius r₁ = 7.400 × 10^6 m to circular orbit radius r₂ = 1.110 × 10^7 m about Earth. Use GM = 3.986 × 10¹⁴ m³ s⁻², take gravitational potential energy zero at infinity, and neglect mass loss and all dissipative forces. Derive the total energy of a circular orbit, calculate the required increase in satellite mechanical energy and the changes in kinetic and potential energy, and explain how a higher orbit can require energy while having lower speed. No transfer-path calculation is required.

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ΔE_total = +2.514 × 10^9 J; ΔU = +5.027 × 10^9 J; ΔK = −2.514 × 10^9 J. The higher circular orbit has lower speed but greater total energy. Gravity provides centripetal force, GMm/r² = mv²/r, so v² = GM/r. Consequently the circular-orbit kinetic energy is K = GMm/(2r), while potential energy is U = −GMm/r. Their sum is E = −GMm/(2r). Initially E₁ = -7.541 × 10^9 J and finally E₂ = -5.027 × 10^9 J; the higher orbit is less tightly bound because its energy is less negative. The required mechanical energy increase is ΔE = E₂ − E₁ = +2.514 × 10^9 J. This is the satellite's net energy change, not a claim about any particular propulsion system's efficiency. The potential-energy increase is ΔU = +5.027 × 10^9 J. The kinetic-energy change is ΔK = −2.514 × 10^9 J, and their sum reproduces the positive total increase. A higher circular orbit has lower orbital speed because v is proportional to r⁻¹ᐟ². The decrease in kinetic energy is smaller than the increase in potential energy, so energy must still be supplied overall.

Energy required for a higher circular orbit · set 4

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