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3 years ago

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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