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Juliette [100K]
3 years ago
9

E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha

t is the asteroid's orbital radius?
Physics
1 answer:
anzhelika [568]3 years ago
6 0

To solve the problem, use Kepler's 3rd law : 

T² = 4π²r³ / GM 

Solved for r : 

r = [GMT² / 4π²]⅓ 

but first covert 6.00 years to seconds : 

6.00years = 6.00years(365days/year)(24.0hours/day)(6... 
= 1.89 x 10^8s 

The radius of the orbit then is : 

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓ 
= 6.23 x 10^11m

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3 years ago
According to Newton's Universal Law of Gravitation, when the distance between two interacting objects doubles, the gravitational
maksim [4K]

Answer:

<em>If the distance doubles, the gravitational force is divided by 4</em>

Explanation:

<u>Newton’s Universal Law of Gravitation </u>

Objects attract each other with a force that is proportional to their masses and inversely proportional to the square of the distance.

\displaystyle F=G{\frac {m_{1}m_{2}}{r^{2}}}

Where:

m1 = mass of object 1

m2 = mass of object 2

r     = distance between the objects' center of masses

G   = gravitational constant: 6.67\cdot 10^{-11}~Nw*m^2/Kg^2

If the distance between the interacting objects doubles to 2r, the new force F' is:

\displaystyle F'=G{\frac {m_{1}m_{2}}{(2r)^{2}}}

Operating:

\displaystyle F'=\frac{1}{4}G{\frac {m_{1}m_{2}}{r^{2}}}

Substituting the original value of F:

\displaystyle F'=\frac{1}{4}F

If the distance doubles, the gravitational force is divided by 4

5 0
3 years ago
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