(a) The acceleration of free fall at the surface of a planet is g and the radius of the planet i

(a) The acceleration of free fall at the surface of a planet...

Key Concepts

Newton's Law of Gravitation

This law states that every two masses attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centers. In the context of orbital motion, it provides the basis for calculating the gravitational force acting on a satellite, using the relation g = GM/R², where g is the surface gravitational acceleration, G is the universal gravitational constant, M is the mass of the planet, and R is the planet's radius.

Centripetal Force in Circular Motion

For an object moving in a circular orbit, the inward (centripetal) force required to keep it in orbit must be provided by gravity. By equating the centripetal force, given by m(4?²r/T²) for a circular orbit, to the gravitational force, one derives a relationship between the orbital radius, the gravitational parameters, and the orbital period.

Orbital Period Derivation

Using the equality between gravitational force and centripetal force, and the definition of orbital speed as the circumference of the orbit divided by the period, one can deduce an expression for the orbital period. For very low orbits, where the orbital radius is approximately equal to the planet's radius, this leads to the simplified formula T = 2??(R/g), linking the period directly to the radius and surface gravitational acceleration.

Kepler's Third Law

Kepler's Third Law relates the square of the orbital period of a satellite to the cube of the semi-major axis of its orbit. Although originally formulated for elliptical orbits, it applies directly to circular orbits in orbital mechanics. The derivation of the orbital period using gravitational and centripetal forces is consistent with Kepler's Third Law, and it serves as a fundamental principle in understanding satellite motion around a planet.

Transcript

00:02 In this problem, we have to show that a satellite orbiting really close to a planet with the surface gravity g and the radius capital r is given by the formula, capital t is equal to 2 pi divided by capital r by g the whole root.

00:27 Now, let's start by considering a satellite that is moving in a really low orbit.

00:34 The radius of the orbit can be assumed to be about the radius, so approximately, the radius of the planet, because it's orbiting at a really small distance above the surface, and the force of, rather, the acceleration can be assumed to be the acceleration on the surface of the planet.

01:07 Now, any object undergoing circular motion must have an acceleration, a centripetal acceleration, directed towards the center.

01:21 And in this case, that will be equal to the gravitational acceleration.

01:26 So a sub c is equal to g, and we can use the formula for the centripetal acceleration.

01:35 That is v squared divided by r, but r is approximately equal to capital r.

01:41 So we'll end up with the equation, we squared is equal to r times g, and that gives we is equal to the square root of g times r.

01:54 Now, we also know that the total distance that the satellite travels is equal to the circumference of the circle.

02:03 That is just equal to 2 pi times capital r.

02:08 And we know that for any object undergoing circular motion, the speed is a constant.

02:18 The speed v is given by the total distance cover, which is c, divided by the time period, capital t.

02:27 So square root of g times r is equal to 2 pi r divided by capital t.

02:39 And that gives capital t is equal to 2 pi times r divided by the square root of g times r.

02:49 So we can go ahead and split the square roots.

02:56 So we'll use square root of g times square root of r.

03:01 And whenever we have x divided by square root of x provided the x squared and zero, it's a positive number, then we can't actually say greater than equal to zero because we can't have a zero in the denominator.

03:17 So this is actually just equal to the square root of x because this is just x to the first power and this is just x to the 1 half power, and that is equal to x to the 1 minus 1 half power, and that is x to the negative 1 half power, which is just equal to, sorry, this is positive 1⁄2, which is equal to the square root of x.

03:44 So that gives t equals 2 pi times the square root of r divided by the square root of g, and we can distribute the root again over the numerator and the denominator to get t equals 2 pi times the square root of r divided by g and our required result is hence shown so for the second part we're given the gravitational acceleration on the surface of a planet is 4 .5 meters per second squared and it has a radius of 3 .4 into 10 to 6 meters and we need to find the time period of a very low orbit.

04:35 So we'll just simply use the formula.

04:37 T equals 2 pi times the square root of r divided by g.

04:46 And we can simply plug in our value, so that's 2 pi times 3 .4 into 10 to the 6th, divided by 4 .5.

04:59 And that gives us 5 ,461 seconds as our answer.

05:16 And we can divide seconds by 60 to get our answer in minutes, and that gives us 91 minutes as our required time period.

05:36 Now for the last part, we're given that time period of another satellite on this planet is 140 minutes, and we need to find the radius of that orbit.

05:54 So what i like to do is first find out the velocity in terms of the radius.

06:09 So we is equal to c divided by capital t, and the circumference is simply 2 pi times the radius of the orbit...

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