Physics, 18.12.2019 05:31 rn2be1980

Modeled after the two satellites problem (slide 12 in lecture13, universal gravity), imagine now if the period of the satellite is exact 88.59 hours, the earth mass is 5.98 x 1024 kg, and the radius of the earth is 3958.8 miles, what is the distance of the satellite from the surface of the earth in miles? use g=6.67x 10 -11 nm2/kg2. your answer could be a large number on the order of tens of thousands, just put in the raw number you get, for example, 12345.67.

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Answer from: mandy9386

R = 6.3456 10⁴ mile

Explanation:

For this exercise we will use Newton's second law where force is gravitational force

F = m a

The satellite is in a circular orbit therefore the acceleration is centripetal

a = v² / r

Where the distance is taken from the center of the Earth

G m M / r² = m v² / r

G M / r = v²

The speed module is constant, let's use the uniform motion relationships, with the length of the circle is

d = 2π r

v = d / t

The time for a full turn is called period (T)

Let's replace

G M / r = (2π r / T)²

r³ = G M T²²2 / 4π²

r = ∛ (G M T² / 4π²)

We have the magnitudes in several types of units

T = 88.59 h (3600 s / 1h) = 3.189 10⁵ s

Re = 6.37 10⁶ m

Let's calculate

r = ∛ (6.67 10⁻¹¹ 5.98 10²⁴ (3,189 10⁵)²/4π²)

r = ∛ (1.027487 10²⁴)

r = 1.0847 10⁸ m

This is the distance from the center of the Earth, the distance you want the surface is

R = r - Re

R = 108.47 10⁶ - 6.37 10⁶

R = 102.1 10⁶ m

Let's reduce to miles

R = 102.1 10⁶ m (1 mile / 1609 m)

R = 6.3456 10⁴ mile

Answer from: Quest

If football was tossed in air without gravity or friction the ball would go into space

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

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