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julsineya [31]
2 years ago
13

The International Space Station (ISS) orbits Earth at an altitude of 4.08 × 105 m above the surface of the planet. At what veloc

ity must the ISS be moving in order to stay in its orbit?
A) 7.91 × 10^3 m/s
B) 3.12 × 10^4 m/s
C) 7.66 × 10^3 m/s
D) 8.17 × 10^3 m/s
Physics
1 answer:
alukav5142 [94]2 years ago
3 0

This question involves the concepts of orbital velocity and orbital radius.

The orbital velocity of ISS must be "7660.25 m/s".

The orbital velocity of the ISS can be given by the following formula:

v=\sqrt{\frac{GM}{R}}

where,

v = orbital velocity = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m

R = 67.86 x 10⁵ m

Therefore,

v=\sqrt{\frac{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}{67.86\ x\ 10^5\ m}}

<u>v = 7660.25 m/s</u>

Learn more about orbital velocity here:

brainly.com/question/541239

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Explanation:

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        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

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X axis

        Tₓ = m a

they relate it is centripetal

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we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

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         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

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