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zhannawk [14.2K]
3 years ago
8

You are explaining to friends why astronauts feel weightless orbiting in the space shuttle, and they respond that they thought g

ravity was just a lot weaker up there. convince them and yourself that it isn't so by calculating how much weaker gravity is at h = 330 km above the earth's surface.
Physics
1 answer:
MissTica3 years ago
8 0
The Earth's radius is 6371 km. So that's our distance from the center when we're on the surface.

The Shuttle astronaut's distance from the center, when s/he's in orbit, is 330 km greater ... that's 6701 km.

The force of gravity is inversely proportional to the distance between the center of the Earth and the center of the astronaut. So, in orbit, it's

(6371/6701)^2 = 90.4 %

of its value on the surface.
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The acceleration of gravity at the surface of Moon is 1.6 m/s2 . A 5.0 kg stone thrown upward on Moon reaches a height of 20 m.
kramer

Answer:

(a) 8 m/s

(b) 5 s

Explanation:

(a)

Using,

V² = U²+2gh ......................... Equation 1

Where V = final velocity, U = Initial velocity, g = acceleration due to gravity on the surface of the moon, h = height reached.

Given: V = 0 m/s ( At it's maximum height), g = -1.6 m/s² ( as its moves against gravity), h = 20 m.

Substitute into equation 1

0 = U²+[2×20×(-1.6)]

-U² = - 64

U² = 64

U = √64

U = 8 m/s.

(b)

V = U +gt.................... Equation 2

Where t = time to reach the maximum height.

Given: V = 0 m/s ( At the maximum height), g = -1.6 m/s² ( Moving against gravity), U = 8 m/s.

Substitute into equation 2

0 = 8+(-1.6t)

-8 = -1.6t

-1.6t = -8

t = -8/-1.6

t = 5 s.

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9.8 \frac{m}{s^2}=\frac{v^2}{6.381 \times 10^6m}   \\  \\ 9.8 \frac{m}{s^2} \times 6.381 \times 10^6m=v^2 \\  \\ v= \sqrt{9.8 \frac{m}{s^2} \times 6.381 \times 10^6m} =24755 \frac{m}{s}
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