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So, I have been given some homework to refresh me for the exam. I am bored so it goes here. Ignore as you please.
(Q1a) We have an illustration of five masses on a line. The first is 5kg. The second (and middle) is 0.1kg and 0.4m to the right. The third (and rightmost) is 10kg 0.6m from the middle mass.
Question is: In the figure given below what are the magnitude and direction of the net gravitational force exerted on the 0.100kg uniform sphere by the other two uniform spheres? The centres of all three spheres are on the same line.
If'n I am not mistaken the force between each pair of speres is given by F = (Gm1m2)/r2. G is the gravitational constant (6.67x10-11N[m2/kg2]), m is the mass of each of the spheres and r is the distance between them. So the force between the first two is F = 0.5G/0.16 = 3.125G = 2.0853125 (which rounds to 2)
Similarly, the force between the middle and rightmost spheres should be G/0.36 = 1.85361111 (rounds to 2).Since I was only given one significant figure to work with and unless I have mucked things up the force experienced by the middle sphere is 2 newtons to the left and 2 to the right and cancels out.
(Q1b) According to Newton's third law, does the 0.100kg sphere exert froces of the same magnitude as your answer to part (a), but opposite direction, on each of the other two spheres?
Short answer: Yes.
Longer answer: To every action (force applied) there is an equal but opposite reaction (equal force applied in the opposite direction) - Newton's Third Law, quoted from Wikipedia.
(Q2a) The mass of Venus is 81.5% that if the earth, and its radius is 94.9% that if the Earth. Compute the acceleration due to gravity on the surface of Venus from these data.
The acceleration due to gravity is g = Gm/R2 with m being the mass of the planet (Venus in this case)
The radius of Earth is 6,378,100 metres so the radius of Venus is 6,052,816.9 metres
The mass of Earth is 5.9742 x 1024 kilograms so the mass of Venus is 4.868973 x 1024 kilograms
And so the answer is that g = 4.868973 x 1024G/(6,052,816.9)2 = 4.868973 x 1024G/4.06801596 x 1013 = 7.99m/s2 (Searching tells me the real answer is more like 8.87)
(Q2b) What is the weight of a 5.00 kg rock on the surface of Venus?
Since weight equals mg, w = 5.00 x 8.87 = 44.35 Newtons
(Q3) The star Rho Cancri is 57 light years from the Earth and has a mass of 0.85 times that of our Sun. A planet has been detected in a circular orbit around Rho Cancri with an orbital radius equal to 0.11 times the radius of the Earth's orbit around the Sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho Cancri?
Orbital speed v is equal to the square root of (Gm)/r, with m being the mass of Rho Cancri and r being the radius of the planet's orbit. The Sun's mass is 1.99 x 1030kg, meaning Rho Cancri has a mass of 1.6915 x 1030. Using the semi-major axis for the 'radius' of Earth's orbit, it is 149,576,887,500m, making the radius of the planet's orbit 16,453,457,625m. And plugging those numbers gets a result for the speed of 6.86018693 x 109 m2/s2.
Okay, so what is the period? We can find that from the expression v = [2(pi)r]/T, with T being the period. So just swap things around T = 15.0695782. I rather strongly suspect that has come out wrong (this is why one should always keep track of units!), although it does look close to the actual period of 14.67 days (then again, the elements I used were slightly different from the actual ones too).
(Q4) The star 70 Virginis is 59 light years from the Earth and has a mass of 1.9 x 1030kg.
(a) A large planet of mass 1.3 x 1028kg is known to orbit this star. The planet is attracted to the star by a force of 3.3 x 1026N, when their centres are seperated by a distance equal to the semi-major axis of the planet's orbit. Calculate this distance (in km).
(b) What is the period (in days) of this planet's orbit?
It is just big numbers, right? For part (a), F, m1, m2 and G are given. r = the square root of (Gm1m2)/F = 70,672,762.7 kilometres
For part (b) I am provided with Kepler's Third Law (Three Laws: Not just for Newton!), T2 = (4[pi]2/GM)a3 (a, by the way, is that pesky semi-major axis again)
So T = 121.340803 days, rounds to 120 days. The real answer: 116 days
Okay, that's that taken care of. Now, I have my real presentation in a couple of hours to prepare for. All calculations performed by google calculator, btw.
~_~
Tricia
