First, some things you should consider:
1) Are you sure you are entering exponential notation correctly intoyour calculator? If you are not sure, see me (office hours?) and I'll check your approach. Here are practice problems on exponential notation.
2) Organization: Start by noting all the information you are given in the problem, and making sure you understand what solution you seek. Then find the equation you need to get you from the known quantities to the unknown solution.
3) Here are some related word problems to practice on (unit conversion, distance, etc).
4) The answers to the problems given below are given at the end of this page. Try doing the problem before you check the answer. If you don't get the right answer, work backwards from the correct answer to check your work and identify where you went wrong.
1) You read a newspaper article saying the angular diameter of the moon will be 31 arcmin across. What is this angle in degrees? In arcsec? Answer
2) Your textbook tells you the distance from the Earth to the moon is 385,500 km. Given what you read in the newspaper (problem #1), what is the physical size of the moon (in km)? Answer
3) At a Stargaze session, you look through the telescope and measure the angular diameter of Jupiter to be 47 arcsec. A person on the Stargaze Staff mentions that tonight, the distance between Earth and Jupiter is 6.287 x 108 km. What is the physical diameter of Jupiter? Answer
1) Venus has an average distance from the sun of 0.623 AU(footnote). How long does it take Venus to complete one orbit around the sun? Answer
2) Mars takes 1.88 years to complete one orbit around the sun. What is its average distance from the sun? Answer
3) Compute the mass of Jupiter given observations of the Galilean moons: the moon Io takes 1.796 days to complete an orbit, and it orbits at an average distance of 423,000 km (units are in days/km/kg, so use G = 4.98x10-10). Answer
1) Compute the force of gravity attracting the moon to the Earth, given that the mass of the moon = 7.35x1022 kg, the mass of Earth = 5.98x1024 kg, and they are separated by 3.44 x 108 m (units are in kg/m so use G=6.67x10-11). Answer
2) Compute the force of gravity attacting a person to the Earth. Assume the person has a mass of 60 kg. The distance from the center to the surface of the Earth is 6,378,000 m (use G=6.67x10-11). Answer
3) Compare the strengths of the two forces. Answer
I.1) You read a newspaper article saying the angular diameter of the moon will be 31 arcmin across. What is this angle in degrees? In arcsec?
(a) We know there are 60 arcmin in 1 degree, so each arcmin is 1/60 of a degree, or 0.01667 degree per arcmin. 31 of these is 0.5167 degrees (mathematically: 31 arcmin x 0.01667 deg/arcmin = 0.5167 deg -- note how the arcmin top & bottom cancel out).(b) We know there are 60 arcsec in 1 arcmin, or 60 arcsec/arcmin. In 31 arcmin, there must be 31 x 60 = 1860 arcsec (mathematically: 31 arcmin x 60 arcsec/arcmin = 1860 arcsec -- note how the arcmin top & bottom cancel out).
I.2) Your textbook tells you the distance from the Earth to the moon is 385,500 km. Given what you read in the newspaper (problem #1), what is the physical size of the moon (in km)?
You know: d = 385,500 km and A = 31 arcmin = 1860 arcsec. You seek: s in km. Use: the small angle formula.Remember that A must be in arcsec for the SAF to work. Rearrange the SAF to isolate s on one side: s = (A x d) / 206265 = (1860 arcsec x 385,500 km) / 206265 arcsec = 3476 km = the physical diameter of the moon [note now arcsec cancels out top and bottom].
I.3) At a Stargaze session, you look through the telescope and measure the angular diameter of Jupiter to be 47 arcsec. A person on the Stargaze Staff mentions that tonight, the distance between Earth and Jupiter is 6.287 x 108 km. What is the physical diameter of Jupiter?
You know: d = 6.287 x 108 km and A = 47 arcsec. You seek: s in km. Use: the small angle formula.Remember that A must be in arcsec for the SAF to work. Rearrange the SAF to isolate s on one side: s = (A x d) / 206265 = (47 arcsec x 6.287 x 108 km) / 206265 arcsec = 1.43 x 105 km = the physical diameter of Saturn [note now arcsec cancels out top and bottom].
II.1) Venus has an average distance from the sun of 0.623 AU. How long does it take Venus to complete one orbit around the sun?
You know: a = 0.623 AU (astronomical units). You seek: the period, P. Use: the simple form of Kepler's Third Law: P2 = a3, where the units are Earth-years and AU.Isolate P on one side of the equation by taking the square root of both sides (the square root of a number squared is the just the number): P = sqrt(a3) = sqrt(0.623 x 0.623 x 0.623) = 0.492 years. The size of an elliptical orbit around the sun is related to the time it takes to complete orbit (the larger the orbit, the longer the time).
II.2) Mars takes 1.88 years to complete one orbit around the sun. What is its average distance from the sun?
You know: P = 1.88 years. You seek: a. Use: the simple form of Kepler's Third Law: P2 = a3, where the units are Earth-years and AU.Isolate a on one side of the equation by taking the cube root of both sides: cubert(P2) = a = cubert(1.88 x 1.88) = cubert(3.53) = 1.52 AU. [This involves taking a cube-root with your calculator -- if you don't know how to do this, try raising 3.53 to the power 0.3333333].
II.3) Compute the mass of Jupiter given observations of the Galilean moons: the moon Io takes 1.796 days to complete an orbit, and it orbits at an average distance of 423,000 km (units are in days/km/kg, so use G = 4.98x10-10).
You know: P = 1.796 days, and a = 423,000 km. You seek: the mass M. Use: the general form of Kepler's Third Law: P2 = (4 x pi2 x a3) / (G x M).Isolate M on one side of the equation by muplitplying both sides by M and dividing both sides by P2, then cancelling top/bottom where appropriate: M = (4 x pi2 x a3) / (G x P2) = (4 x pi2 x (423,000)3) / (4.98 x 10-10 x (1.796)2) = 1.86 x 1027 kg.
III.1) Compute the force of gravity attracting the moon to the Earth, given that the mass of the moon = 7.35x1022 kg, the mass of Earth = 5.98x1024 kg, and they are separated by 3.44 x 108 m (units are in kg/m so use G=6.67x10-11).
You know: m = 7.35x1022 kg and M = 5.98x1024 kg, and r = 3.44 x 108 m. You seek: the gravitational force F. Use: Newton's Law of Gravity: F = -G x M x m / r2.No algebraic rearrangement necessary, since F is already isolated: F = -G x M x m / r2 = - (6.67x10-11) x (5.98x1024) x (7.35x1022) / (3.44x108)2 = 2.48 x 1020 Newtons.
III.2) Compute the force of gravity attacting a person to the Earth. Assume the person has a mass of 60 kg. The distance from the center to the surface of the Earth is 6,378,000 m (use G=6.67x10-11).
You know: m = 60 kg and M = 5.98x1024 kg, and r = 6,378,000 m. You seek: the gravitational force F. Use: Newton's Law of Gravity: F = -G x M x m / r2.No algebraic rearrangement necessary, since F is already isolated: F = -G x M x m / r2 = - (6.67x10-11) x (5.98x1024) x (60) / (6,378,000)2 = 588 Newtons.
III.3) Compare the strengths of the two forces.
You know: F(moon) = 2.48 x 1020 Newtons and F(you) = 588 Newtons. A useful way to compare the magnitudes of two quantities is to take a ratio of the two values.F(moon) / F(you) = 2.48 x 1020 Newtons / 588 Newtons = 4.22 x 1017. That is, the gravitational force exerted by the Earth on the moon is about 42 quadrillion times larger than the gravitational force exerted by the Earth on you. [Gravity works best on big masses, even though in this case the distance is larger].
Footnote: I
miscopied this number from the back of the book ... it should be
0.723 AU. If you check your answer in the back of the book, it will
disagree. I'll fix this after people are done with the exam. Thanks
for finding my mistake, Bethany!