Little little g

This problem involves four different parts which you can either discuss, just think about or analyse with various levels of detail.

 

The effective weight of an object at any place on earth mainly depends on three things:

 

1. The gravitational pull of the earth on the object.

2. The centripetal acceleration on the object caused by the rotation of the earth on its axis.

3. The mass of the object.

 

The gravitational acceleration is typically quoted as $g$ is $9.80665\mathrm{ms}^{-2}$ and the weight $W$ of an object as $W=mg$.

 

Part 1: The figure quoted in this question for $g$ assumes that the earth is a sphere. Newton's law of gravitation says that the gravitational acceleration felt at a distance $R$ from the centre of a uniform sphere is given by

$$

g =\frac{GM}{R^2}\;,\quad G = 6.67300\times 10^{-11} \mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}.$$

What radius does this imply for the earth?

 

Part 2: In 1968 the Olympic Games were held in Mexico city, at an altitude of $2240\mathrm{m}$ above sea level. At these games, Bob Beamon jumped a staggering $8\mathrm{m}$ $90\mathrm{cm}$ in the long jump, smashing the previous record by $55\mathrm{cm}$. This record survived until 1991 when it was broken by a small amount in Tokyo (altitude $17\mathrm{m}$), by Mike Powell.

Do you think that the unusually high altitude of Mexico City contributed to the longevity of Bob Beamon's record? Back up your thoughts with an analysis.  

 

Part 3: Is the rotational effect of any significance on your weight? Do as much analysis as seems necessary to determine 'significance'.

 

 

Part 4: Is there anything else that might have a tiny effect on your weight?