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tino4ka555 [31]
3 years ago
10

Two satellites, one in geosynchronous orbit (T = 24 hrs) and one with a period of 12 hrs, are orbiting Earth. How many times lar

ger than the radius of Earth is the distance between the orbits of the two satellites.
Physics
1 answer:
VashaNatasha [74]3 years ago
6 0

The distance between the orbits of two satellites is 7.97 m.

Explanation:

Johannes Kepler was the first to propose three laws for the planetary motion. According to him, the orbits in which planets are rotating are elliptical in nature and Sun is at the focus of the ellipse. Also the area of sweeping is same.

So based on these three assumptions, Kepler postulated three laws. One among them is Kepler's third law of planetary motion. According to the third law, the square of the time taken by a planet to cover a specified region is directly proportional to the cube of the major elliptical axis or the radius of the ellipse.

So, T^{2} = r^{3}

Thus, for the geosynchornous satellite, as the time taken is 24 hours, then the radius or the major axis of this satellite is

(24)^{2}= r^{3} \\(2*2*2*3)^{2} = r^{3}\\r = \sqrt[\frac{2}{3} ]{2*2*2*3} =(2)^{2} * (6)^{\frac{2}{3} } =13.21 m

Similarly, for the another satellite orbiting in time period of 12 hours, the major axis of this satellite is

(12)^{2}= r^{3} \\(2*2*3)^{2} = r^{3}\\r = \sqrt[\frac{2}{3} ]{12}  =5.24 m

So, the difference between the two radius will give the distance between the two orbits, 13.21-5.24 = 7.97 m.

So the distance between the orbits of two satellites is 7.97 m.

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DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0.12×3×10⁸ m/s

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a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s

Time taken to reach 12% of light speed is 3673469.39 seconds

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m

The distance it would have to travel is 6.61×10¹⁴ m

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3 years ago
A truck with 0.420-m-radius tires travels at 32.0 m/s. what is the angular velocity of the rotating tires in radians per second?
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Angular velocity of the rotating tires can be calculated using the formula,

v=ω×r

Here, v is the velocity of the tires = 32 m/s

r is the radius of the tires= 0.42 m

ω is the angular velocity

Substituting the values we get,

32=ω×0.42

ω= 32/0.42 = 76.19 rad/s

= 76.19×\frac{1}{2\pi} *60 revolution per min

=728 rpm

Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.

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