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3 years ago

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Which formulas have been correctly rearranged to solve for radius? Check all that apply. r = GM central/v^2 r =fcm/v^2 r =ac/v^2

r =vt/2pi r =act/pi

2 answers:

jek_recluse [69]3 years ago

7 0

The orbital radius is: $r=\frac{GM}{v^2}$

Explanation:

The problem is asking to find the radius of the orbit of a satellite around a planet, given the orbital speed of the satellite.

For a satellite in orbit around a planet, the gravitational force provides the required centripetal force to keep it in circular motion, therefore we can write:

$\frac{GMm}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

M is the mass of the planet

m is the mass of the satellite

r is the radius of the orbit

v is the speed of the satellite

Re-arranging the equation, we find:

$\frac{GM}{r}=v^2\\r=\frac{GM}{v^2}$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

11Alexandr11 [23.1K]3 years ago

6 0

Answer:

a and d

Explanation:

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Answer:

2.15 m

Explanation:

Newton's Law of Universal Gravitation:

$\displaystyle F_g = G \frac{m_1 m_2}{r^2}$

$\displaystyle F_g$ is the gravitational force of attraction

$\displaystyle G$ is the universal gravitational constant

$m_1$ and $m_2$ are the two masses of the two objects

$\displaystyle r$ is the distance between the centers of the two objects.

List the known values:

$\displaystyle F_g = 3.47 \cdot 10^-^8 \ \text{N}$
$\displaystyle G = 6.673 \cdot 10^-^1^1 \ \frac{Nm^2}{kg^2}$
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Plug these values into the equation:

$\displaystyle 3.47 \cdot 10^-^8 \ \text{N} = 6.673 \cdot 10^-^1^1 \ \frac{Nm^2}{kg^2} \frac{(44.8 \ \text{kg})(53.9 \ \text{kg})}{r^2}$

Notice that the units $\displaystyle \text{N}$ , $\displaystyle \text{kg}^2$ , and $\displaystyle \text{m}$ cancel out. We are left with the unit $\displaystyle \text{m}$ for radius r.

Get rid of the units to make the problem easier to read.

$\displaystyle 3.47 \cdot 10^-^8= 6.673 \cdot 10^-^1^1 \ \frac{(44.8)(53.9) \ }{r^2}$

Multiply the masses together.

$\displaystyle 3.47 \cdot 10^-^8= 6.673 \cdot 10^-^1^1 \ \frac{(2414.72)}{r^2}$

Multiply the gravitational constant and the masses together.

$\displaystyle 3.47 \cdot 10^-^8= \frac{1.61134265\cdot 10^-^7}{r^2}$

Solve for $r^2$ by dividing both sides by 3.47 * 10^(-8) and moving $r^2$ to the left.

$\displaystyle r^2 = \frac{1.61134265\cdot 10^-^7}{3.47\cdot 10^-^8}$
$r^2 = 4.64363876$

Take the square root of both sides.

$r=2.154910383$

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