Gravitation

1. (a) Distance travelled by satellite in one period
= 2. pi.(3.59x10^7 + 6.14x10^6) m = 2.64 x 10^8 m

Speed of satellite = square-root[GM/R]
where G is the Gravitational constant = 6.67 x 10^-11 m^3/kg.s^2
M is the mass of the earth = 5.97 x 10^24 kg
R = (3.59x10^7 + 6.14x10^6) m = 4.2 x 10^7 m

Period = distance/speed = 2.64x10^8/(GM/R)

(b) g = GM/R^2
Weight of astronaut = 50g

The asronaut feels weightlessness because he moves with the same speed as the satellite. There is no relative motion between the two.
He same experience is found if the both astronaut and satellite fall freely un der gravity.

2. Acceleration due to gravity on earth surface, g = GM/R^2
where G is the Gravitational constant and M is the mass of earth
Acceleration due to gravity at 2R, g' = GM/(2R)^2 = GM/4R^2 = g/4
Hence, weight of the 1 kg mass at 2R is 1/4 of its weight on earth surface.

3. Since mR(2.pi/T)^2 = GmM/R^2
where m is the mass of the moon
R is the moon-earth distance
M is the mass of earth (= 5.97 x 10^24 kg)
G is the Gravitational constant (= 6.67 x 10^-11 m^3/kg.s^2)
T is the period of the moon (= 27.3 x 24 x 3600 s)

i.e. R^3 = GMT^2/(4.pi^2)

4. (a) The distance to the centre of the earth from the equator is longer than that from the two poles. Hence, the apparent weight decreases when moving towards the equator from the north pole, and increases again when further moving to the south pole.

(b) Centripetal force at equator = 1 x (6.37x10^6) x (2.pi/(24 x 3600))^2 N = 0.0337 N
Hence, apparent weight at equator = (g - 0.0337) N
where g is the acceleration due to gravity at the poles
Hence, %age difference = [g - (g - 0.0337)]/g x 100% = 0.34%
[Take g = 9.81 m/s^2]


2015-07-03 10:36:36 補充：
sorry, a typo in the last sentence in Q1. It should be
Period = distance/speed = 2.64x10^8/square-root(GM/R)

2015-07-04 00:04:56 補充：
Q:where did you find the mass of earth?
A: It can be found from any physics textbook or from the web.

2015-07-04 00:06:06 補充：
Q:what does it mean by 'no relative motion'?
A: They both under the same acceleration and have the same speed all the time.

2015-07-04 00:10:01 補充：
Q:why apparent weight should be using g-0.0337?
A: The mass is 1 kg, hence earth attraction on the mass = 1 x g = g
Because the mass moves in circular motion along with earth rotation at the equator, part of the attractive force needs to provide for centripetal acceleration.

2015-07-04 00:13:48 補充：
(cont'd)...
Since centripetaql force = 1 x 0.0337 N
hence, net attractive force pulling the object downward = (g - 0.0337) N

hello where did you find the mass of earth?
and what does it mean by 'no relative motion'?

2015-07-03 20:19:29 補充：
why apparent weight should be using g-0.0337? (g- centripetal force)