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Pepsi [2]
3 years ago
6

If the same satellite orbited at the same velocity around a planet with three times the force of gravity, the orbit radius would

be (multiplied, divided) by ___________.
Physics
1 answer:
belka [17]3 years ago
5 0

Answer:

the orbit radius would be divided by 3

Explanation:

In order to orbit the planet, the centripetal force of satellite must balance the gravitational force between planet and satellite.

centripetal Force = Gravitational Force

(Ms)(V)²/r = (G)(Ms)(Mₓ)/r²

V² = GMₓ/r

where,

V = velocity of satellite

G = Gravitational Constant

Mₓ = Mass of Planet

r = orbit radius

but, G = gr²/Mₓ

Therefore,

V² = (Mₓ/r)(gr²/Mₓ)

V = √gr   --------------------- equation (1)

where,

g = force of gravity

Now, for change in force of gravity:

V' = √g'r'  

where,

g' = 3g

V' = V

Therefore,

V = √3gr'   ---------------- equation (2)

Comparing equation (1) and equation (2), we get:

√gr = √3gr'

gr = 3gr'

r = 3r'

r' = r/3

<u>Hence, the orbit radius would be divided by 3.</u>

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A woman (m = 45kg) is kneeling on the shoulders of a man (m = 70kg) in pair figure-skating while both are traveling at 1.5 m/s.
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Both the man and the woman will be experiencing the same impulse, due to Newton's third law of motion.

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What is the ideal banking angle (in degrees) for a gentle turn of 2.00 km radius on a highway with a 125 km/h speed limit (about
nikitadnepr [17]

An "ideal" banking angle assumes no friction is required to keep a car on the road as it turns. Let <em>θ</em> denote the banking angle, and consult the attached free-body diagram for a car making the turn. There are only 2 relevant forces acting on the car,

• the normal force with magnitude <em>n</em>

• the car's weight with magnitude <em>w</em>

and the net force points toward the center of the circle made by the turn, with centripetal acceleration

<em>a</em> = (125 km/h)² / (2.00 km) = 7812.5 km/h² ≈ 0.603 m/s²

Split up the forces into components acting perpendicular (⟂) and parallel (//) to the banked curve, so that by Newton's second law,

∑ <em>F</em> (⟂) = <em>N</em> + <em>W</em> (⟂) = <em>m</em> <em>a</em> (⟂)

and

∑ <em>F</em> (//) = <em>W</em> (//) = <em>m a</em> (//)

Let the direction of <em>N</em> be the positive perpendicular axis, and down the incline and toward the center of the circle the positive parallel axis. The net force vector and acceleration both make an angle <em>θ</em> with the banked curve, and <em>W</em> makes the same angle with the negative perpendicular axis, so that the equations above reduce to

<em>N</em> - <em>m g</em> cos(<em>θ</em>) = <em>m</em> <em>a</em> sin(<em>θ</em>)

and

<em>m g</em> sin(<em>θ</em>) = <em>m a</em> cos(<em>θ</em>)

The second equation is all we need at this point to find the ideal <em>θ</em>. The mass <em>m</em> cancels out, and we can solve for <em>θ</em> to get

tan(<em>θ</em>) = <em>a</em>/<em>g</em> ≈ (0.603 m/s²) / (9.80 m/s²) ≈ 0.0615

→   <em>θ</em> ≈ 3.52°

8 0
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