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## Main Question or Discussion Point

An unmanned spacecraft moves in a circular orbit with speed

Now the question is, after calculating using conservation of Energy Method,

[K+U][tex]_{initial}[/tex]=[K+U][tex]_{final}[/tex].

the landing speed of the spacecraft is smaller than the escape velocity. why is that so? any one?

Thanks

*v*around the moon,observing the lunar surface at an orbital radius of*R*=1790 km.The period of the orbit is*T*.Unfortunately, an electrical fault causes an on board thruster to fire, reducing the speed of the spacecraft by [tex]\Delta[/tex]*v*=5 m/s. If nothing is done to correct its orbits. with what speed (in km/h) will the spacecraft crash into the lunar surface? Mass of moon=7.35[tex]\times[/tex]10[tex]^{22}[/tex] kg and Radius of Moon = 1740 km. Universal gravitational constant, G = 6.673[tex]\times[/tex]10[tex]^{-11}[/tex] Nm[tex]^{2}[/tex]/kg[tex]^{2}[/tex].Now the question is, after calculating using conservation of Energy Method,

[K+U][tex]_{initial}[/tex]=[K+U][tex]_{final}[/tex].

the landing speed of the spacecraft is smaller than the escape velocity. why is that so? any one?

*If the spacecraft orbiting without the electrical fault, is that speed suppose to be greater than,or lesser than then escape velocity in order for it to maintain at that orbit.*Thanks

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