Physics, 03.11.2020 17:00 cisneros3

Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The altitude of a geosynchronous orbit is

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Answer from: hdhdjjdj80HDjdjdj

Complete Question

Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. These are called geosynchronous orbits. The altitude of a geosynchronous orbit is 3.58×107m(≈22,000miles).

Part A

What is the period of a satellite in a geosynchronous orbit?

Part B

Find the value of g at this altitude.

Part A

the period of a satellite in a geosynchronous orbit is 24 hours

Part B

the value of g at this altitude is $g = 0.224 \ m/s^2$

Explanation:

From the question we are told that

The altitude of the geosynchronous orbit is $r = 3.58* 10^7 \ m$

Generally 24 hr make up a day , which means that in 24 hours the earth does a complete rotation about its axis

Now from the question we are told that communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotate.it then means that the communications satellites has the same time period as the earth given that it is in a fixed position with respect to the earth

Generally the value of g at this altitude is mathematically represented as

$g = \frac{G * M }{(R + r )^2}$

Here G is the gravitational constant with value $G = 6.67 *10^{-11} \ N \cdot m^2 \cdot kg^2$

also M is the mass of the earth with value $M = 5.97 *10^{24} \ kg$

and R is the radius of the earth with value $R = 6.38 *10^{6} \ m$

$g = \frac{6.67*10^{-11} * 5.97*10^{24} }{( 6.38*10^{6} + 3.58*10^{7} )^2}$

=> $g = 0.224 \ m/s^2$

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

Answer from: Quest

Ireally just need to get finished with precal! i’m really sorry i have to do this! hope this !

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