Your Friendly Physics Tutor

[dustybray]

Hi,

I’m having a really difficult time in physics, and I need some help on this homework assignment.

Here are the questions and the work I’ve done. I really don’t know how to approach them, and at this point, I’m just kinda throwing equations at them haphazardly…


5. A 175 kg weather surveillance satellite of is launched into orbit with an altitude of
250 km. The mass of the Earth is 5.97 x 1024 kg. The radius of the Earth is
6.38 x 103 km.


a. Calculate the tangential velocity required for this orbit.

mg = mv^2 / r

g = v^2 / r

v = sqrt(gr) = sqrt( 9.8m/s^2 * (6.38*10^6 m + 250000m) )

v = 8060.6m/s

b. Calculate the period of this orbit in minutes.

T^2 = (4PI^2 / GM[E]) r^3

T^2 = (4PI^2 / (6.67 * 10^-11) * 5.97 x 1024 kg) * (6.38 x 106 m + 250000m)^3

T = 3.2*10^28s/rev

3.2*10^28s/rev * (1min / 60s) = 5.33*10^26 min/rev ??

c. What is the weight of the satellite?

mg = G( Mm / r^2 ) I can’t use this formula, there are two m’s…


v^2 = G( m/r )

m = rv^2 / G

m = ( (6.38*10^6 m + 250000m) * (8060.6m/s)^2 ) / (6.67*10^-11)

This isn’t right…


9. In an Atwood machine, one block has a mass of 500g and the other a mass of 460g.
The pulley, which is mounted in horizontal frictionless bearings, has a radius of
5.0cm. When released from rest, the heavier block is observed to fall 75cm in 5.0s
with no slippage of the cord on the pulley.

d = v*t + (1/2)*a*t^2

d-v*t = (1/2)*a*t^2

a = 2*( d -v * t)/ t^2 )

a = 2*( ( .75m -0 * 5s) / (5s)^2 ) = .06m/s^2

a. Calculate the acceleration of the two blocks.

a[M] = -.06m/s^2
a[m] = .06m/s^2

b. Calculate the tension in the cord supporting the heavier block.

T[1] = mg+ma = (.46kg) (9.8m/s^2) + (.46kg) (.06m/s^2) = 4.54N

c. Calculate the tension in the cord supporting the lighter block. Note the answer to b and c is not the same.

T[2] – T[1] = Mg – mg – Ma – ma

T[2] – 4.54N = (.5kg) (9.8m/s^2) – (.46kg) * (9.8m/s^2) – (.5kg) * (.06m/s^2) – (.46kg) * (.06m/s^2)

T[2] – 4.54N = .33N

T[2] = 4.87N

d. Calculate the angular acceleration of the pulley.

α = a / r = .06m/s^2 / .05m = 1.2rad/s^2

e. Calculate the moment of inertia of the pulley.

T[2] – T[1] = I (a / r^2)

I = (T[2] – T[1]) / (a / r^2)

I = (.33N) / (.06m/s^2 / (.05m)^2) = .014 kg*m^2


Any help would be greatly appreciated.

Thanks,

dusty…….

 

[Tide]

5. $v^2/r$ is the acceleration and you know the mass. So the weight is ...?? :)

 

[quicknote]

Dear dusty,

I can understand your frustration with these problems. It seems like you are on the right track with your equations and calculations, but there are a few things that need to be corrected.

For problem 5, part b, your calculation for the period of the orbit is incorrect. The correct formula to use is T = 2π * √(r^3 / GM[E]), which gives a period of approximately 5.6 hours. For part c, the weight of the satellite can be found using the formula W = mg, where m is the mass of the satellite (175 kg) and g is the acceleration due to gravity at the satellite's altitude (approximately 8.9 m/s^2). This gives a weight of approximately 1550 N.

For problem 9, part c, the tension in the cord supporting the lighter block should be equal to the weight of the lighter block (0.46 kg) plus the force due to its acceleration (0.46 kg * 0.06 m/s^2), which gives a total tension of 4.74 N.

I hope this helps you in approaching these problems and understanding the concepts better. If you have any further questions, feel free to reach out for assistance. Keep up the good work in your physics studies!

Sincerely,