Spaceship trajectory

For entering a circular orbit around another planet, a spacecraft must change its trajectory upon arrival; otherwise it will fly past the planet. If this trajectory change is done with a rocket engine alone, the fuel required for this can contribute significantly to the cost of the mission. However, if the spacecraft’s trajectory carries it through the atmosphere of the planet, then the resulting drag can significantly reduce the amount of fuel required. Consider a 2,500 kg spacecraft to be sent to the planet Neptune, arriving at a speed of 30 km/s. How much margin of error would be allowed in designing a trajectory through the outer atmosphere of Neptune which would save a significant amount of fuel?

The problem was asked in UPh Competition. I can't understand how drag can reduce energy. If viscous drag is there, it would require more energy to accelerate!!What is margin of error?

Easy Math Editor

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Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$