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VLD [36.1K]
3 years ago
15

A new planet is discovered that has twice the Earth's mass and twice the Earth's radius. On the surface of this new planet a per

son who weights 500 N on Earth would experience a gravitational force of ______.
(A) 125 N
(B) 250 N
(C) 500 N
(D)) 1000 N
(E) 2000 N
Physics
1 answer:
Margarita [4]3 years ago
8 0

Answer:

option B

Explanation:

Radius of new planet,R' = 2R

Mass of the earth, M' = 2 M

R and M is the Radius and Mass of the earth.

Weight of the person on earth = 500 N

Weight of the person in the new planet = ?

We know acceleration due to gravity is calculated by using formula

   g = \dfrac{GM}{R^2}

now, acceleration due to gravity on the new planet

   g' = \dfrac{GM'}{R'^2}

   g' = \dfrac{G(2M)}{(2R)^2}

   g' =\dfrac{1}{2} \dfrac{GM}{R^2}

   g' =\dfrac{g}{2}

here acceleration due to gravity is half in new planet, so weight will also be half on the new planet.

Weight on the new plane is equal to \dfrac{500}{2} = 250 N

correct answer is option B

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I attached a diagram that shows how this force aligns with the force of gravity.
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A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri
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3 years ago
A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
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Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

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r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

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3 years ago
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KEf = KEi + PEi

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Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h

So:
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</span>1/2 m * vf² =  1/2 m * vi² + m * a * h
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Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²

Potential energy depends on mass (m), acceleration (a), and height (h):
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Divide both sides by m:
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1/2 * 11.53² = 9.8 * h
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66.47 = 9.8 * h
h = 66.47 / 9.8
h = 6.78 m
3 0
3 years ago
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