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Newton’s law of universal gravitation

Motion in the heavens

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2) The table below contains data about the sizes and periods of planetary orbits around the sun.

A) use the data to calculate the orbital speed of each planet and fill in the last column. You can assume that the shapes of the orbits are circles. The orbits are really ellipses as you have learned, but the flattening is very slight, and using circles simplifies the calculation.

To calculate orbital speed, use the formula V = 2 pi r / T

V = orbital speed

r = orbital radius

T = orbital period (time for the planet to go around the Sun once)

Pi= 3.14      !! your answer will be in km/years so convert to km/s,

1 year = 365.2 x 24 x 60 x 60 s  (check on line, how many days in a year, I am not sure exactly)

T in seconds = T in years x 3.16 E7

 

 

 

B) How do the orbital speeds change from Mercury to Pluto ?

 

C) Try to show B using Mathematics: The gravitational force acting on the planet keeps them to fly away. The expression is: Fg= G M m / r²  .

M is the mass of the Sun

m is the mass of the planet

 r is the distance Sun-planet

This force is the centripetal force keeping the planets in orbit around the Sun. You know that the centripetal force Fc = ________. (use m the mass , v the orbital speed and r )

Using the fact that Fc = Fg, solve for v = _________.

So as r increases, V _________     Is this consistent with your answer in A ?

3) Newton found the universal law of gravitation. The attractive gravitational force between 2 masses m1 and m2 located at a distance r one from the other is : Fg = G m1 m2 /r².

G is a constant. (6.67 E^-11 Nm² /kg² ). The force between 2 masses (m1 and m2) is proportional to their product but inversely proportional to the square of the distance  between their centers. Newton didn’t find G. It was Cavendish who found G by doing a very precise experiment and was able to “weight the Earth”.

 

By substituting any changes in any of the variables into this equation, we can predict how the others change. For example, we can see how the force changes if we know how either or both of the masses change, or how the distance between their centers changes.

Example1: suppose that one mass somehow is doubled. Then substituting 2m, for m1, in the equation gives:

F_new = G 2 m1 m2 /r² = 2 (G m1m2/ r² ) = 2 Fold  So F_new = 2 F_old

So we see the force doubles also.

 

Example2: Suppose that that the distance of separation is doubled. Then substituting 2r for r in the equation gives :

F_new = G m1 m2 (2r)² = G m1 m2 /4r² = 1/4 (G m1 m2 /r2) = 1/4 (F_old)

And we see the force is only 1/4 as much. F_new = 1/4 F_old.

 

Following example 1 and example 2 answer these questions:

A) If both masses are doubles, what happens to the force?

 

B) If the masses are not changed, but the distance of separation is reduced to 1/2 the original distance, what happens to the force ?

 

C) If the masses are not changes, but the distance of separation is reduced to 1/4 the original distance, what happens to the force  ?

 

D) If both masses are doubled, and the distance of separation is doubled, show what happens to the force.

 

E) If one of the masses is doubled, the other remains unchanged, and the distance of separation is tripled, show what happens to the force

 

F) Consider a pair of binary stars that pull on each other with a certain force. Would the force be larger or smaller if the mass of each star were three times as great when their distance apart is three times as far ? Show what the new force will be compared to the first one.

4) The gravitational force Fg keeps the planets in orbits around the Sun.  So the gravitational force is a centipede force Fc. The planet keeps falling toward the sun. Suppose you have a planet of mass m located at a distance r from the Sun. The mass of the sun is M.

Remember the expression for any centripetal force ?

Fc = m V² /r   .

 (m is the planet’s mass, V its speed around the Sun and r the distance from the Sun)

You saw previously that:  V = 2 pi r / T. V is the orbital speed and T the orbital period.

 

A) substitute V into Fc. You have now: Fc = ___________  (was our lab, by the way) hint: Fc should only include m, T, r and pi.

 

B) Using the expression for Fc found in A and using Fg = G m M / r² and Fc = Fg show that:

 

(4 pi² ) r³ / T² = G M

 

this means that the mass of the sun can be found from the period and radius of a  planet’s orbit ! isn't that something ?    The mass of the planet can be found only if it has a satellite orbiting it.

 

C) for extra credits: (on a separate piece of paper, show your work)

 Using: (4 pi² ) r³ / T² = G M

Can you show Kepler’s Third law ? (take 2 planets orbiting the Sun 1 and 2. don’t use numbers)

 

(r1/r2)³ = (T1/T2) ²

The square of the ratio of the periods of any 2 planets is equal to the cube if the ratio of their distance from the Sun.

(Remember, you found it using the Kepler’s worksheet)

 

 

 

5) Kepler’s law says: (r1/r2)³ = (T1/T2) ²

 

r1 is the distance from the Sun of planet 1, r2 is the distance from the Sun of planet 2, T1 is the orbital period of planet 1 and T2 is the orbital period of planet 2.

Jupiter is 5.2 times farther from the Sun than Earth is. Find Jupiter’s orbital period in Earth years. (hint: T1 = 1 year, T2 = ?, r2 = 5.2 r1)

 

6)Tom has a mass of 70kg (m1) and Sally has a mass of 50kg (m2). Tom and sally are standing 20m ®  apart on the dance floor. Sally looks up and sees Tom. She feels an attraction. If the attraction is gravitational (Fg) , find its size. Assume that both and Tom and Sally can be replaced by spherical masses. IS the attraction gravitational then?

(use Fg = G m1m2/r2, with G = 6.6710-11 )

 

7) Assume that you have a mass of 50kg and Earth has a mass of 5.97 E24 kg. The radius of Earth is about 6370km.

A) What is the force of gravitational attraction between you and the Earth ?

hint: Use: Fg = G m1 m2 /r²     G = 6.67 E –11

B) You also know that  weight (force of gravity) = mass x 9.8.  (because F = ma)

Compute the weight of the mass of 50kg (w = mg). Do you get the same answer as in A?

C) Convert the weight (or Fg) in pound.

You can use: 9.8N = 2.2 lb on Earth

8) From 7) you know that, at the surface of the Earth, Fg = mg with g = 9.8. You also know that Fg = GmM/R². R is the radius of the Earth

(6400km) and M the mass of the Earth. The weight and Fg are the same force.

A) calculate the mass of the Earth M

hint: Fg = G m1 M /r² = m1 g, with g = 9.8m/s/s, M is the mass of the Earth, G = 6.6710^-11solve for M.

 

B) Calculate the average density of Earth in kg/m3 then in g /cm3

hint: volume of sphere = 4/3 pi r³, density = Mass/ Volume of course

 

C) the value found should be close to 5.5 g/cm3. Density of iron (core of Earth)  is 7.87 g/cm3 and density of granite (crust) is 2.75g/cm3. So is it surprising?

 

9) using (4 pi² ) r³ / T² = G M answer A and B

A) Use one line of the table of 2) to find the mass of the Sun. Check in Wiki if you are right. T is seconds, r in meters, G = 6.67 E –11 solve for M.

 

B) Optional: Mimas, one of Saturn’s moons, has an orbital radius of about 1.87 E8 m and an orbital period of about 23h. Find the mass of Saturn. Check if you are right.

Remember convert hours to seconds

10) optional, advanced Physics

Use Newton’s law of universal gravitation to find an equation where x is equal to an object's distance from the Earth's center, and y is its acceleration due to gravity. Use a graphing calculator to graph this equation, using 6400-6600 km as the range for x. Trace along and find y:

A) at sea level, 6400km

B) on top of Mt Everest, 6410km

C) In a typical satellite orbit, 6500km

D) in a much bigger orbit, 6600km

11) extra credits (if you show your work neatly)

 

If you go up by only 100 km (from the ground where R = 6370km), Gravity is weaker by how many percent ?

Hint: New Gravity g = GM / r²,  M is the mass of the Earth (you don’t need it), r is the total distance from the center of Earth = 100 km + 6370km and old gravity go = GM / R² with R = 6370km

% difference = |new - old| / old

 

12)

Wherever you are, your mass m stays constant (same amount of matter) but the force due to gravity varies. (not taking in account relativity effects)

A) what is your mass in kg on Earth (If it is a secret, use mine, m = 65kg ) ? (1kg = 2.2 lb)

B) What is you weight in Newton on Earth (W_earth = mg of course, g = 9.8m/s/s)

C) On Jupiter , what is your mass ?

D) On Jupiter, what is your weight (in Newton) USE: gjupiter = 22.88 m/s/s ?

W_jupiter = ___________N

E) So your weight on Jupiter = _________ times your weight on Earth. We can’t stand more than about 10 times our weight. What will happen to your bones on Jupiter (suppose you hold your breath)?

F) The acceleration due to gravity on Jupiter is therefore ___________ times the acceleration on Earth.

g_jupiter = ____ g_earth

 

 

SORRY , THERE IS MORE/ CLIK HERE (this is so much Fun !)