Calculating Orbital Periods in Saturn's Rings

In summary, the distance between Venus and the sun is 1.08x10^8 km and the distance between Earth and the sun is 1.50x10^8 km. The length of the Venusian year can be estimated using Newton's gravitational force law, assuming circular orbits, and is equal to the orbital velocity divided by the radius. The rings of Saturn are composed of chunks of ice and have an inner radius of 73,000 km and an outer radius of 170,000 km. The mass of Saturn is 5.69x10^26 kg. Using the same formula, the period of orbiting chunks of ice at the inner and outer radii can be found. The mass of the grav
  • #1
solo
5
0
i have no idea how to do this :(

Venus is an average distance of 1.08x10^8 km from the sun. Estimate the length of the Venutian year given that the Eearth is 1.50x10^8 from the sun on the average?

answer in years.
one more :(

the rings of saturn are composed of chunks of ice that orbit the planet. the inner radius of teh rings is 73,000 km, while the outer radius is 170,000 km. The mass of saturn is 5.69x10^26 kg.

a) find period of orbiting chunk of ice at inner r
b) find period of orbiting chunk of ice at outer r

im thikning I am going to use the formula: F = m1m2 / r^2 but I am not sure how to approach it

help would be appreciated


thanks dudes
 
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  • #2
For all of these problems, start with Newton's gravitational force law (which you gave, except you left out the gravitational constant G); assume that the orbits are circular, so that the gravitational force is a centripetal force. Use the centripetal force to find the orbital velocity, and with that and the radius, you can find the period.
 
  • #3
F = m v^2 / r is centripetal force right

and thanks for the help
 
  • #4
Yes, that's the right formula for centripetal force.
 
  • #5
coudl you please clarify how to use the radius and orbital velocity to find the period?

my texctbook doesn't give a good explanation?

thanks

F = m v^2 / r

so it would be G m m(e) / r^2 = m v^2 / r

what would m(e) be
 
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  • #6
coudl you please clarify how to use the radius and orbital velocity to find the period?

If you know the distance something travels (e.g., the circumference of the orbit), and its speed, you can find out how long it takes to travel that far.


F = m v^2 / r
so it would be G m m(e) / r^2 = m v^2 / r
what would m(e) be

In this context, it's the mass of the gravitating body about which the body of mass m is orbiting.
 
  • #7
yes thank you it is finally clear

G ( 5.69 x 1-^26) / (170,000,000)^2 = v^2/r

finally

i got it :D
 

1. What is an easy gravitational problem?

An easy gravitational problem is a physics problem that involves calculating the force of gravity between two objects. This can include finding the gravitational force between two masses, determining the acceleration due to gravity, or solving for the distance between two objects based on their masses and gravitational force.

2. How do you calculate the force of gravity?

The force of gravity between two objects is calculated using the equation F = G * (m1 * m2)/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

3. What is the gravitational constant?

The gravitational constant (G) is a fundamental constant in physics that represents the strength of the force of gravity between two objects. It has a value of approximately 6.67 x 10^-11 N * m^2/kg^2.

4. Can you provide an example of an easy gravitational problem?

An example of an easy gravitational problem would be calculating the force of gravity between a 5 kg mass and a 10 kg mass that are 2 meters apart. Plugging these values into the equation F = G * (m1 * m2)/r^2, we get F = 6.67 x 10^-11 N * m^2/kg^2 * (5 kg * 10 kg)/(2 m)^2 = 1.67 x 10^-10 N.

5. How does distance affect gravitational force?

The force of gravity between two objects is inversely proportional to the square of the distance between them. This means that as the distance between two objects increases, the force of gravity between them decreases. This relationship can be seen in the equation F = G * (m1 * m2)/r^2, where r is the distance between the two objects.

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