1 the international space station m 45 x 105 kg orbits at an...

1- The International Space Station (m = 4.5 x 105 kg) orbits at an altitude of 415 km above the Earth’s surface. (a) Does the orbital speed of the station depend on its mass? Explain. (b) Find the speed of the station in its orbit (c) How much time does it take for the station to complete one orbit of the Earth?

Transcript

0:00 Hello.

00:01 First we can solve part a of this question.

00:04 In this question, the space station is in stable orbit around the earth.

00:08 Therefore we can write fc is equal to fg, where fc is the centripetal force and fg is the gravitational force accepted by the earth on the space station.

00:18 Now fc can be written as m v squared divided by r, where m is the mass of the space station, v is the orbital speed of the space station and r is the radius of orbit of the space station.

00:30 It can be written as equal to fg can be written as g m divided by r square.

00:38 Here g is gravitational constant and capital m is the mass of the earth.

00:43 Simplifying this equation, the equation for orbital speed can be written as v is equal to root of gm divided by r.

00:53 Here g and m are constants, therefore the orbital speed v is inversely proportional to the square of.

01:00 Root of r, where r is the radius of orbit of the space station.

01:05 Therefore we can conclude that the orbital speed v does not depend upon the mass m of the space station.

01:20 Now in part b we can write r is equal to capital r plus h.

01:28 Here r is the radius of the earth and h is the height of the space station with respect to the earth's surface.

01:34 The value of r is equal to 6 ,000 370 km, it can be written as equal to 6 ,370 into 10 raised to 3 meter.

01:47 H is given to be equal to 415 km, it can be written as equal to 415 into 10 raise to 3 meter.

01:58 We can substitute the values here so it equals 6 ,370 into 10 raised to 3 meter plus 415 into 10 raised to 3 meter plus 415 into 10 raised to 3 meter...

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