from the average value of T 2 /a 3 in V o R f R e a T=500 T 2 /a 3 7.91 6.39E+06
ID: 2020836 • Letter: F
Question
from the average value of T2/a3 in
Vo
Rf
Re
a
T=500
T2/a3
7.91
6.39E+06
6.32E+06
6.35E+06
5050
4.016
0.0055
8.00
6.61E+06
6.37E+06
6.49E+06
5200
4.166
0.0185
8.2
7.34E+06
6.37E+06
6.85E+06
5650
4.660
0.0708
8.8
1.03E+07
6.37E+06
8.32E+06
7600
6.942
0.2361
9.2
1.32E+07
6.37E+06
9.79E+06
9700
9.611
0.3488
9.6
1.77E+07
6.37E+06
1.20E+07
13150
14.41
0.4721
10
2.52E+07
6.37E+06
1.58E+07
19750
9.89
0.5959
10.2
3.13E+07
6.37E+06
1.88E+07
25650
35.0
0.6630
a)Calculate the Mass of the earth
b)Compare it to (5.97*1024).Find the difference
c) Explain how the mass of the sun could be determined from informayion about the planetary orbits
Vo
Rf
Re
a
T=500
T2/a3
7.91
6.39E+06
6.32E+06
6.35E+06
5050
4.016
0.0055
8.00
6.61E+06
6.37E+06
6.49E+06
5200
4.166
0.0185
8.2
7.34E+06
6.37E+06
6.85E+06
5650
4.660
0.0708
8.8
1.03E+07
6.37E+06
8.32E+06
7600
6.942
0.2361
9.2
1.32E+07
6.37E+06
9.79E+06
9700
9.611
0.3488
9.6
1.77E+07
6.37E+06
1.20E+07
13150
14.41
0.4721
10
2.52E+07
6.37E+06
1.58E+07
19750
9.89
0.5959
10.2
3.13E+07
6.37E+06
1.88E+07
25650
35.0
0.6630
Explanation / Answer
(a) For earth, orbital velocity vo = 7.91 km/s Average value of T^2/a^3 = 11.1 x10^-20 s^2/ m^3 [ from the table ] T^2 /a^3 = 4^2 / GMe 11.1 x10^-14 = 4^2 / GMe Mass of the earth, Me = 4^2/ (11.1 x10^-14* 6.67x10^-11) = 5.33 x 10^24 Kg (b) Difference = 5.97x10^24 - 5.33 x 10^24 = 0.64 x 10^24 kg (c) Mass of the sun, Ms = (4^2 /G) (a^3/T^2)Related Questions
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