Circular motion + gravitation
2011-11-15 2:11 am
圖片參考：http://imgcld.yimg.com/8/n/HA06571321/o/701111140063613873401440.jpg
a) Find the angular speed ω of the Earth and the centripetal accelration of the Earth round th Sun.
b) Find the gravitational force acting on the Earth due to the Sun
c) Given that the universal gravitation constant G is 6.67 X10(-11) Nm(2)kg(-2)
find the mass M of the sun
d) Find the gravitational field strength due to the sun at the orbit taken by the Earth. Hence explain briefly why a person on Earth does not feel weightless as in the case an astronaut in a space shuttle moving round the Earth.
e) The planet Jupiter thakes 11.9 Earth year to complete one orbit of the Sun.
Calculate the radius of Jupiter's orbit.
回答 (3)
2011-11-15 4:14 am
✔ 最佳答案
(a) Since period of revolution of the earth is 365 days (= 3.15x10^7 s)Angular speed w = 2.pi/3.15x10^7 s^-1 = 1.99 x 10^-7 s^-1
Centripetal acceleration = 1.5x10^11 x (1.99x10&-7)^2 m/s^2 = 5.968 x 10^-3 m/s^2
(b) Gravitational force = (6x10^24) x (5.968x10^-3) N = 3.58 x 10^22 N
(c) Since gravitational attraction = GM(6x10^24)/(1.5x10^11)^2
where M is the mass of the sun
Hence, GM(6x10^24)/(1.5x10^11)^2 = 3.58 x 10^22
M = 2 x 10^30 kg
(d) Gravitational field strength = 3.58x10^22//6x10^24 N/kg = 5.97 x 10^-3 N/kg
The field strength of the sun is so weak as compared with that given the the earth. Hence, a person on earth is primarily attracted by earth gravity.
(e) Using Kepler's Law,
11.9^2 = [R/(1.5x10^11)]^3
where R is the radius of Jupiter's orbit
solve for R gives R = 7.82 x 10^11 m
2011-11-16 2:11 pm
The Kepler's Law which he used is the Kepler's Third Law. It can be proved easily.
Let M, m, R and w be the mass of the Sun, the mass of the Jupiter,the orbital radius of the Jupiter and the angular welocity of the Jupiter respectively. As the centripetal force of the Jupiter is provided form the gravitational force of the Sun, we have
GMm/R^2 = mRw^2
GM/R^3 = w^2
GM/R^3 = (2π/T)^2 (where T is the orbital period of the Jupiter)
T^2/R^3 = 4π^2/GM = constant
In this case, it is known that the orbital period of the Earth is 1 year and its orbital radius is 1.5x10^11 m. By putting T=1 and R=1.5x10^11, the constant 4π^2/GM is obtained, i.e. [1/(1.5x10^11)]^3. Hence, we have
11.9^2 / R^3 = [1/(1.5x10^11)]^3
11.9^2 = [R/(1.5x10^11)]^3
R = 7.82 x 10^11 m
The Kepler's Third Law is useful for calculating the period or the radius of a planet when the period or the orbital radius of that planet is given.
Be careful of using Kepler's Third Law as the constant depends on the mass of the star.
Let M, m, R and w be the mass of the Sun, the mass of the Jupiter,the orbital radius of the Jupiter and the angular welocity of the Jupiter respectively. As the centripetal force of the Jupiter is provided form the gravitational force of the Sun, we have
GMm/R^2 = mRw^2
GM/R^3 = w^2
GM/R^3 = (2π/T)^2 (where T is the orbital period of the Jupiter)
T^2/R^3 = 4π^2/GM = constant
In this case, it is known that the orbital period of the Earth is 1 year and its orbital radius is 1.5x10^11 m. By putting T=1 and R=1.5x10^11, the constant 4π^2/GM is obtained, i.e. [1/(1.5x10^11)]^3. Hence, we have
11.9^2 / R^3 = [1/(1.5x10^11)]^3
11.9^2 = [R/(1.5x10^11)]^3
R = 7.82 x 10^11 m
The Kepler's Third Law is useful for calculating the period or the radius of a planet when the period or the orbital radius of that planet is given.
Be careful of using Kepler's Third Law as the constant depends on the mass of the star.
參考: you should select 天同's solution as the best answer
2011-11-15 4:40 am
a)ω=2π/T =2π÷(365×24×60×60) =1.99×10-7 rads-1a=rω2 =1.5×1011×(1.99×10-7)2 =5.95×10-3 ms-2b)F=ma =(6×1024) ×(5.95×10-3) =3.571022Nc)F=GMm/R2A=GM/R23.571022=6.67 X10-11×M÷(1.5×1011)2M=1.21×1055kgd)E=F/m =5.95×10-3ms-2∵E<<<<g=10 ms-2So a person on Earth does not feel weightlesse)r13/T12=r23/T22(1.5×1011)3/T12= r23/(11.9 T1)2r2=7.82×1011m
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