Radius of the Earth

[Kyorisu]

Quick physics help here people. Another dull physics prac that doesn't count towards my S.A.C.E. (south australian certificate of education) and thus counts only towards my school grade (a,b,c,d) anywhoo I need an easy way to calculate the radius of the earth. We know the radius isn't a constant but I don't have to be that specific. This is so I can calculate the Earth's mass.

So far I've devised and ran the experiment to find accelertaion due to gravity and due to some error (stopwatches) I got 9.5ms per second per second. Anywhoo I can either rerun that test or use those numbers without losing marks.

G is the gravitational constant, M is the mass of the Earth and r is the radius of the Earth.

M = ar2/G, where a=9.8m/sec*2, r=6.4 x 10*6m, and G=6.67 x 10*-11m*3/(kg sec*2).

M = 9.8 x (6.7 x 10*6)*2 /6.7 x 10*-11 = 6.0 x 10*24 kg

note *= to the power of. In this instance I used 9.8/sec*2 unless I feel like reruning the test I'll change it to 9.5.

Now the only problem is I cannot assume anything or use any knowledge unless proven so in essence I need a quick way to get r.

I could and are using the idea of the moon being attracted to the Earth which doesn't rely on using acceleration due to gravity or the radius of the earth but I thought my chosen method would tie in better as we must prove acceleration due to gravity and hence find the Earth's mass.

 

[AntiAnto]

http://spiff.rit.edu/classes/phys240/homework/earth_radius/earth_rad.html

http://vamana.space-india.org/phase_I_feb28.htm

http://teacher.pas.rochester.edu/phy121/LectureNotes/Chapter14/Chapter14.html#Heading3

14.2. The gravitational constant G

The strength of the gravitational force depends on the value of G. The value of the gravitational constant can be determined using the Cavendish apparatus. Two small lead spheres of mass m are connected to the end of a rod of length L which is suspended from it midpoint by a fine fiber, forming a torsion balance. Two large lead spheres, each of mass M, are placed in the location indicated in Figure 14.3. The lead spheres will attract each other, exerting a torque on the rod. In the equilibrium position the gravitational torque is just balanced by the torque exerted by the twisted fiber. The torque exerted by the twisted wire is given by

Figure 14.3. The Cavendish Apparatus.

The torque exerted by the gravitational force is given by

where R is the equilibrium distance between the center of the large and the small spheres. If the system is in equilibrium, the net torque acting on the rod is zero. Thus

All of a sudden the large spheres are rotated to a new position (position B in Figure 14.3). The net torque acting on the twisted fiber is now not equal to zero, and the system will start to oscillate. The period of oscillation is related to the rotational inertia and the torsion constant [kappa]

The angle between the two equilibrium positions is measured to be 2[theta]. This, combined with the measured torsion constant, is sufficient to determine the torque [tau] acting on the torsion balance due to the gravitational force. Measurements show that G = 6.67 x 10-11 Nm2/kg2.

 

[DEATH eVADER]

diameter of the Earth is 12,756Km, that would make the radius 6378Km

 

[Kyorisu]

I heard someone talking about that method in class. Thanks for the link.

Funny though I've been googling for quite a while and didn't find anything to suffice.

Like I said ealier I can't just state the diameter I must first have a proof for said Diameter before I can use it.

 

[DEATH eVADER]

http://gpc.edu/~pgore/Earth&Space/GPS/planets.html

 

[Kyorisu]

A proof that can be done in person ie myself. We know that acceleration due to gravity is 9.8 meters per second per second, yet I had to devise a prac for that by dropping a ball from a 2 story building and using g = 2d/t*2 (g=gravity, d=distance and t=time)

Thanks for the info though I can probably squeeze it in as a proven fact anyway for bonus marks. Anywhoo I'd better write some of this stuff down before bed. I have a hour and 40 minute physics lesson tommorow morning.

 

[kirovman]

Off the top of my head, 6400km.

That number has been so useful to me in everyday life.

Now the only problem is I cannot assume anything or use any knowledge unless proven so in essence I need a quick way to get r.

So can you assume the value of G? Or the formula? Because they have not been 'proven', there's only strong evidence to suggest they are valid.

 

[Kyorisu]

I can use the gravitational constant that is the only number we may freely use. I should have mentioned that. Otherwise proving that gives me even more bonus marks.