In a shuttle craft of mass m = 2100 kg, Captain Janeway orbits a planet of...

In a shuttle craft of mass m = 2100 kg, Captain Janeway orbits a planet of mass M = 5.98

Expert Solution

This problem is full of redundant information. It asks to determine the semimajor axis of the elliptical orbit after the maneuver. So, let us start from the final question and work backwards.

The equation that relates the semimajor axis of the elliptical orbit "a", the velocity of the craft "v", the mass of the planet and the radius (the radial position), is the "Vis-viva equation".

(1)

Here G is the gravitational constant.

Since the maneuver is brief, we know that the radius could not have changed much. Then, it remains constant. The problem gives us the new velocity and the mass of the planet.

The problem is now solved. We have just to replace values and solve for "a".

Replacing r=6.7*10e6m, M=5.98*10e24Kg, V=7.29*10e3m/s (this is the new velocity), G=6.67384*10e11 m3/(Kg*s2) in the latter formula is:

a=6.07*10e6 m.

The semimajor axis of the elliptical orbit is close to six million meters. This value makes sense, since the initial radius is 6.7 million meters (when the semimajor axis equals the radius) and the variation in speed is not big, then the variation in the semimajor axis cannot be too big.

All the other given information is redundant and unnecessary.

genius_generous answered 6 months ago

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