A satellite is in a circular orbit 200km above the surface of the earth. If its

Question

A satellite is in a circular orbit 200km above the surface of the earth. If its acceleration is 9.2m/s2, what is its speed? 2. What is the minimum value of the coefficient of static friction if a car is able to go around a flat curve of radius 80m at a speed of 80km/hr? 3. A 0.5kg ball on a string is being swung is a vertical circle of radius 0.7m. a) If the tension in the string is 2N at the top of the circle, what is the ball A satellite is in a circular orbit 200km above the surface of the earth. If its acceleration is 9.2m/s2, what is its speed? 2. What is the minimum value of the coefficient of static friction if a car is able to go around a flat curve of radius 80m at a speed of 80km/hr? 3. A 0.5kg ball on a string is being swung is a vertical circle of radius 0.7m. a) If the tension in the string is 2N at the top of the circle, what is the ball

Explanation / Answer

1)That is the radial acceleration

so a = v^2/r

therefore v = sqrt(r*a) = sqrt((6.38x10^6 + 200x10^3)*9.2) = 7780m/s

2)n= V^2/rg=.629

3)a)3.7m/s

b)10.7N

4)rom the law of gravitation GMm/r^2 = m*a

we see that a is proportional to 1/r^2

At Earth's surface (R = 6.38x10^6m) the radial acceleration is g which = 9.80m/s^2

Now set up the ratio

9.80/(1/R^2) = 2/(1/R + h)^2

or 9.80*R^2 = 2*(R + h)^2

rewriting

(R + h)/R = sqrt(9.80/2)

so (R + h) = 2.2136*R

so h = 1.2136*R = 1.2136*6.38x10^6 = 7.74x10^6m = 7740km above the Earth's surface

5)F1 = GMo*m/x^2

F2 = G*2Mo*m/(R-x)^2
They are equal in magnitude when:

GMo*m/x^2 = F1 = F2 = G*2Mo*m/(R-x)^2
1/x^2 = 2/(R-x)^2
(R-x)^2 = 2*x^2
R^2 - 2Rx + x^2 = 2*x^2
0 = x^2 + 2Rx - R^2

=> x = (-2R

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