Circular and Satellite Motion Legacy Problem #24 Guided Solution
Problem*
Hercules is hoping to put a baseball in orbit by throwing it horizontally (tangent to the Earth) from the top of Mount Newton - 97 km above Earth's surface. With what speed must he throw the ball in order to put it into orbit? (GIVEN: MEarth = 5.98 x 1024 kg; REarth = 6.37 x 106 m)
Audio Guided Solution
If Hercules is to put his ball into orbit he must throw it with a sufficient enough speed tangent to the earth such that the rate at which it falls towards the earth matches the rate at which the earth curves. Newton has figured out for us that this speed is dependent upon the radius of its orbit and the mass of the earth according to the equation v equal the square root of big G times the mass of the earth divided by the radius of the orbit. So if we are to determine the speed of our baseball what we have to do is know two things the mass of the earth which is given here and the radius of orbit which is not given but which we can calculate. Now if you think about the radius of the orbit of this baseball it is a little bit bigger than the radius of the earth. If you were to draw a picture of the earth representing it as a circle and then to think about the location of the ball relative to the earth it is slightly above the earth's surface and it is going to orbit in a circle as well a circle which is concentric with earth's circle but is simply a little bit larger by an amount of 97 kilometers. So if you wish to calculate the radius of orbit you have to take the radius of the earth and add to it the distance that the satellite is or the baseball above earth's surface. You have to take the 6.37 times 10 to the 6 meters and add to it the height of the baseball which is 97 kilometers and not a great unit for adding to meters. So first take the 97 kilometers and convert it to meters by multiplying by 1000 and then add that value onto the 6.37 times 10 to the 6. This is the radius of orbit. Its value is 6.467 times 10 to the 6 meters. That goes into the denominator inside of the radical sign and in the numerator is big G times the mass of the earth where big G is 6.673 times 10 to the negative 11 and the mass of the earth is stated here. When you take the square root of the whole thing you get about 7,855. We can round that to three significant digits and the answer would be 7.86 times 10 to the third meters per second.
Solution
7.86 x 103 m/s
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities in an organized manner. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(\descriptive{m}{m,mass} = 61.7\unit{kg}\), \(\descriptive{v}{v,velocity} = 18.5 \unit{\meter\per\second}\), \(\descriptive{R}{R,radius} = 30.9\unit{m}\), \(F_\text{norm} = \colorbox{gray}{Unknown}\).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
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