Suppose a spacecraft orbits the moon in a very low, circular...

Suppose a spacecraft orbits the moon in a very low, circular orbit, just a few hundred meters above the lunar surface. The moon has a diameter of $3500 \mathrm{km},$ and the free-fall acceleration at the surface is $1.6 \mathrm{m} / \mathrm{s}^{2}$. How much time does it take for the spacecraft to complete one orbit? A. 38 min B. 76 min C. 110 min D. 220 min

Transcript

00:01 Hello students.

00:02 Now in this question they have given a spacecraft which revolves the moon in a very low in a circular orbit just a few hundred meters above the lunar surface and that moon has the diameter of 3 ,500 kilometers and free fall acceleration at the surface is 1 .6 meter per second square and they are asking us to calculate the time does it takes for the spacecraft to complete one orbit.

00:34 They're asking us to calculate time.

00:37 The time it takes for the spacecraft to complete one orbit around the moon can be calculated using the formula.

00:44 Time t is equal to 2 pi u r by v where t is the time of the orbit and r is the radius of the orbit and v is the velocity of the spacecraft.

01:03 Since the spacecraft, since the spacecar is the time of the orbit, is a very very low circular orbit just a few hundred meters above the lunar surface we can assume that its distance from the center of the moon is approximately equals to the radius of the moon therefore we can take v as the thousand five hundred kilometers by two which is equal to thousand five zero kilometer then the velocity of the spacecraft can be calculated using the centipital acceleration formula that is a is equal to v square by r here a is the centipital acceleration and v is the velocity of the spacecraft and r is the radius of the orbit at the surface of the moon the free fall acceleration is given as 1 .6 meter per second square which is given then setting to the equilibrium equal this equation we will get b is equal to root of a into r then root of a is 1 .6 meter per second square into 1750 kilometer then by putting these values we will get v is equal to 1607 meter per second now we can calculate the period of the orbit using the formula we have here t is equal to 2 pi u r by v then t is equal to 2 into pi into r is we have 1 7 5 0 kilometer by 1607 meter per second then we have by converting kilometers to meters and simplifying we will get t is equal to 2 into 5 into 1 .75 into 10 to the power of 6 meter divided by 1607 meter per second...

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