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maxonik [38]
3 years ago
6

The International Space Station (ISS) orbits Earth in a nearly circular orbit that is 345 km above Earth's surface.

Physics
1 answer:
Tems11 [23]3 years ago
5 0

Answer:

1) v = 7.70 10³ m/s , 2) F = 115 N and 3)    (F/W)% = 90.2%

Explanation:

1) To solve the problem let's use Newton's second law where force is gravitational force and acceleration is centripetal

    F = ma.

    F = G m M / r²

    a = v² / r

    G m M / r² = m v² / r

    G M / r = v²

Let's look for the distance is the distance from the surface of the has to the station 345 103 m plus the radius of the Earth

    r = Re + 345 103

    r = 6.37 10⁶ + 3.45 10⁵

    r = 6.715 10⁶ m

Let's calculate the speed

   v = √ (6.67 10⁻¹¹ 5.98 10²⁴ / 6,715 10⁶) = √ (59,399 10⁶)

   v = 7.70 10³ m/s

The speed module is constant, so we can use the uniform motion relationships

   v = d / t

The distance is the length of the circle

   d = 2π r

   d = 2π 6.715 106

   d = 42.2 10⁶ m

Let's calculate the time

   t = d / v

   t = 42.2 10⁶ / 7.70 10³

   t = 5.48 10³ s

2) Let's use the universal gravitation equation

   F = G m M / r²

   F = 6.67 10⁻¹¹ 13.0  5.98 10²⁴ /(6.715 10⁶)²

   F = 11.5 10¹ N

   F = 115 N

3) in this for we are asked the relationship is out with the weight of the body on earth

   F / W = F / mg

   F / W = 115 / (13.0  9.8)

   F / W = 0.902

  F / W% = 90.2%

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-- Traveling at 40 mph for 1 hour, the distance covered is 40 miles.

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