# A set of practice problems - Gravity and law of gravitation

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1. (a) Calculate the mass of the earth given the acceleration of gravity at the north pole is 9.830 m/s^2 and the radius of the earth is 6371 km from pole to pole. (b) Compare this with the accepted value of 5.979 x 10^24 kg.

2. (a) Calculate the ratio r^3/T^2 for the moon's orbit about the earth, where r is the average radius and T is the period. Express the answer in SI units. (b) Calculate the same ratio for any moon of Jupiter. (c) Why are the answers not equal? (NOTE: do not use Newton's expression for this quantity, just use the raw numbers for period and radius, which you can easily find. Since you live in the age of the internet, I do not have to give you all the data required for homework. Spending 20 seconds with a search engine gives lots of web sites such as http://nssdc.gsfc.nasa.gov/planetary/factsheet/joviansatfact.html)

3. One of Jupiter's moons, Callisto, has a period of 16.69 days, and an orbital radius of 1.88 x 10^9 m. Find the mass of Jupiter from these data, and compare with the accepted mass of Jupiter (1.90 x 10^(27) kg).

4. Find the acceleration of gravity at an altitude above the earth's surface equal to 2 times the radius of the earth. (This is then 3 earth radii from the center of the earth.) You should not need to use the values for the earth's radius or mass, nor do you need G, the gravitational constant.

5. A geosynchronous satellite is in a circular orbit above the equator, with a radius of 6.63 times the earth's radius. If the satellite weighs 4.50 tons at the earth's surface, what is the force of gravity (in tons) on the satellite when it is in orbit? (You don't need to know values for the earth's radius or mass. Just use the fact that the force of gravity is inversely proportional to the square of the distance.)

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-1. (a) Calculate the mass of the earth given the acceleration of gravity at the north pole is 9.830 m/s^2 and the radius of the earth is 6371 km from pole to pole. (b) Compare this with the accepted value of 5.979 x 10^24 kg.

Consider a mass m on the surface of earth at the North Pole. According to the inverse square law of force between two masses, the force is given by F = G m*M/r2

Where M is the mass of the earth and r is its radius. G is the Universal gravitational constant.

At the same time, the force acting on a body of mass m subjected to an acceleration a is given by, F = m * a

We can equate both the equations as they are describing the same force acting on the same body.

So, m a = G m*M/r2

Or, M = a*r2/G

G = 6.67300 × 10-11 m3 kg-1 s-2

 M = 9.830*[6371*103]2 /6.67300 × 10-11

= 5.9792622662970178330585943353814*10+24 Kg

As you can see this is exactly ...

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