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(III) $(a)$ Show that if a satellite orbits very near the surface of aplanet with period $T,$ the density $(=$ mass per unit volume) ofthe planet is $\rho=m / V=3 \pi / G T^{2} .$ (b) Estimate the densityof the Earth, given that a satellite near the surface orbits with aperiod of 85 min. Approximate the Earth as a uniform sphere.

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a) $\frac{3 \pi}{G T^{2}}$b) $5.4 \times 10^{3} k g / m^{3}$

Physics 101 Mechanics

Chapter 6

Gravitation and Newton's Synthesis

Physics Basics

Newton's Laws of Motion

Applying Newton's Laws

Gravitation

Cornell University

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

Lectures

03:43

In physics, dynamics is the branch of physics concerned with the study of forces and their effect on matter, commonly in the context of motion. In everyday usage, "dynamics" usually refers to a set of laws that describe the motion of bodies under the action of a system of forces. The motion of a body is described by its position and its velocity as the time value varies. The science of dynamics can be subdivided into, Dynamics of a rigid body, which deals with the motion of a rigid body in the frame of reference where it is considered to be a rigid body. Dynamics of a continuum, which deals with the motion of a continuous system, in the frame of reference where the system is considered to be a continuum.

03:55

In physics, orbital motion is the motion of an object around another object, which is often a star or planet. Orbital motion is affected by the gravity of the central object, as well as by the resistance of deep space (which is negligible at the distances of most orbits in the Solar System).

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(III) $(a)$ Show that if a…

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(III) ($a$) Show that if a…

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A spherical planet has uni…

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Show that the minimum peri…

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A satellite is in a circul…

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If the period of revolutio…

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A satellite of mass $300 \…

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A satellite of mass $m,$ o…

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A satellite of mass 5500 k…

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A 600-kg satellite is in a…

Oh, here we need the formula for a planet and then we need the density of Earth s so we can say that the force of gravity would be the only force on any satellite, and this would be equal to the mass times these centripetal acceleration. So we can say that this So we can say that the gravitational, constant times the mass of the satellite times the mass of Earth divided by R squared will be equal to M V squared over r. And so we can say that the velocity is going to be equal to the square of G times the mass of the earth divided by our and this would be for the satellite. Now, we know that by definition this will equal to pi r over the period tee, and this is for any object under uniform circular motion. We know that em over r cubed is simply going to be equal to for pi squared over g t squared. So essentially, we have said that the mass we can actually to find this mass of the earth. So here the mass of the earth, divided by r cubed, we've essentially manipulated this and said. That created a density function essentially and this is equaling four pi squared over g t squared. And so here we know that the density is going to be equal to the mass divided by for over three pi r cubed, given that were modelling the earth as a perfect sphere. And we can say that this is going to equal three over rather three over for pie times four pi squared over Gede t squared. So essentially we're plugging in for this m over r cubed function. So this term right here is being substituted with this and so we can finally say that after thes cancel out there, we have that Ah, the density is going to be equal to three pie over g t squared. And that's how we find the jet the density of a planet. Given that we only have ah ah satellite in orbit. So the period of the satellite. So we can say that in this case the period of the satellite is inversely proportional to the density of the plan. And this would be your answer for party and for party for Earth. We're simply going to solve it. So for Earth. We can say that the density that the density of earth, it'll be three pie and then 6.67 times 10 to the negative 11th a new meters squared per kilogram squared, gravitational constant and then multiplied by 85 minutes. Or we can say 60 seconds per minute and quantity squared, and we find that the density of earth is going to be equal to 5.4 times 10 to the third kilograms per cubic meter. So this is the final answer for part B. That is the end of the solution. Thank you for watching.

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