Circular and Satellite Motion Legacy Problem #18 Guided Solution
Problem*
Determine the force of gravitational attraction between the Earth and the sun. Their masses are 5.98 x 1024 kg and 1.99 x 1030 kg, respectively. The average distance separating the Earth and the sun is 1.50 x 1011 m. Determine the force of gravitational attraction between the Earth and the sun.
Audio Guided Solution
Newton's Law of Universal Gravitation states that all objects with mass attract with a force which is directly proportional to the product of their masses and inversely proportional to the square of their separation distance. Applied here to the situation of the Earth and the Sun, we can calculate the force of gravitational attraction using the equation F-grav equal big G, where big G is a proportionality constant sometimes called the universal gravitation constant. Big G multiplied by the mass of the Earth multiplied by the mass of the Sun divided by the square of the separation distance. Now if we plug in for the mass of the Earth 5.98 times 10 to the 24th, for the mass of the Sun 1.99 times 10 to the 30th, and for the distance separating them 1.50 times 10 to the 11th, and then use as a gravitational constant 6.673 times 10 to the negative 11th Newtons meter squared per kilogram squared, we will get the force of gravitational attraction. It comes out here to be 3.53 times 10 to the 22nd Newtons. Now if you got an answer that went something like 3.53 times 10 to the something other than 22nd Newton, then perhaps you are doing it right, only using your calculator incorrectly. Particularly when it comes to scientific notation numbers, it is important that you know how to enter them correctly. If we take a number like 5.98 times 10 to the 24th, the way I would enter that into my calculator is I would type in 5.98, and then push the button E, which means exponent base 10, and then I would type 24th. Your calculator would likely read something like 5.98 big E 24, or 5.98 and then some spaces 24, which means 5.98 times 10 to the 24th. Entering it instead as 5.98 multiplied by 10, and then E to the 24th would enter a number which is 10 times the actual intended number.
Solution
3.53 x 1022 N
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.
Read About It!
Get more information on the topic of Circular and Satellite Motion at The Physics Classroom Tutorial.