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SVEN [57.7K]
3 years ago
5

Use Kepler’s laws and the period of the Moon (27.4 d) to determine the period of an artificial satellite orbiting very near the

Earth’s surface.
Physics
2 answers:
miv72 [106K]3 years ago
7 0

As according to Kepler 's law

T =(4π²r³/ Gm)^1/2

here r= distance  from earth center to satellite = 6400km = 6400000m

G = earth's gravitational constant= 6.67×10^-11

m = mass of earth= 5.98 ×10^24

so T =[ { 4× (3.1416)²×(6400000)³}/ {(6.67×10^-11)×( 5.98 ×10^24)} ]^1/2

T= 5133 sec


natka813 [3]3 years ago
7 0

Answer:

T_2 = 1.40 hours

Explanation:

As per kepler's law of time period we know that

square of time period is proportional to the cube of the distance

so here we can say

\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}

so we know that

for moon

T_1 = 27.4 days

r_1 = 384,400 km

r_2 = 6370 km

now from above formula we have

\frac{27.4^2}{T_2^2} = \frac{384,400^3}{6370^3}

T_2 = 0.058 days

T_2 = 1.40 hours

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A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 4.4 cm from the axis of rotation. (a) Calcul
romanna [79]

Answer:

a) a =0.53 m/s²

b) μ=0.054

c) μ = 0.068

Explanation:

a) If we assume that the turntable is rotating at a constant speed, the only force acting on the seed parallel to the surface, which keeps it  from following a straight line trajectory, is the centripetal force.

So, we can apply Newton's 2nd Law to the seed in this way:

Fnet = m*a = m*ac = m*ω²*r

We have the value of the angular speed, ω, in rev/min, so it is advisable to convert it to rad/sec, as follows:

ω = 33 rev/min*(1 min/60 sec)*(2*π rad/ 1 rev) = 11/10*π rad/sec

So, replacing in (1), we can solve for ac, as follows:

ac = ω²*r = (11/10)²*π²*0.044 m = 0.53 m/s²

b) Now, the centripetal force that we found above, is not a new type of force, it must be a force that explains the behavior of the seed.

As the seed does not slip, the only force acting  on it parallel to the surface, is the static  friction force, which has a maximum value, as follows:

Ff = μ*N

As there is no movement in the vertical direction, this means that the normal force must be equal and opposite to Fg, so we can write the expression for Ff as follows:

Ff = μ*m*g

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Ff  = Fc ⇒ μ*m*g = m*ac

⇒ μ*g = ac

Solving for μ:

μ = ac/g = 0.53 m/s² / 9.8 m/s² = 0.054

c) During the acceleration period, added to the centripetal acceleration, as the angular speed is not constant, we will have also an angular acceleration, γ , which we can get as follows:

γ = Δω/Δt = (11/10)*π / 0.37 s = 9.34 rad/sec²

By definition of angular acceleration, there exists a fixed  relationship between the angular acceleration and the tangential acceleration (same as the one between angular and tangential speed), as follows:

at = γ*r = 9.34 rad/sec²*0.044 m = 0.41 m/s

When the turntable has reached to its maximum angular velocity, it will have also the maximum value of the centripetal acceleration, which we have just found out.

So, the magnitude of the total acceleration (at the moment of maximum acceleration) as they are perpendicular each  other) , is given by the following expression:

a = √(ac)²+(at)² = 0.67 m/s²

Now, as friction force opposes to the relative movement between surfaces (the seed and the turntable), it shall be larger than the product of the mass times the total acceleration, acting along  the same action line, so we can say:

Ffmin = μ*m*g = m*a

⇒ μmin = a/g = 0.67 m/s²/9.8 m/s² = 0.068

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