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Gravity, Kepler's law

by nns91
Tags: gravity, kepler
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nns91
#1
Mar22-09, 02:34 PM
P: 301
1. The problem statement, all variables and given/known data
1. If the mass of a satellite is doubled, the radius of its orbit can remain constant if the speed of the satellite:
a. increases by a factor of 8
b. increases by a factor of 2
c. does not change
d. is reduced by a factor of 8
e. is reduced by a factor of 2

2. Astronauts orbiting in a satellite 300km above the surface of the earth feel weightless. Why ? Is the force of gravity exerted by the earth on them negligible at this height ?

3. Five equal masses M are equally spaced on the arc of a semicircle of radius R. m is located at the center of curvature of the arc. (a) If M is 3kg,m is 2kg, and R is 10cm, what is the force on m due to the 5 masses. If m is removed, what is the gravitational field at the center of curvature of the arc ?

2. Relevant equations

F= -Gm1m2/ r^2, T^2= 4pi^2r^3/GM(s)

3. The attempt at a solution

1. I chose e but I am wrong. What would be the appropriate way to find out the answer of this question ??

2. I said that the distance r^2 is really big so F is negligible at this height. I am wrong again. How should I explain this ??

3. So I did this problem wrong. Is it true that 4 of M are canceled because they are opposite ? So the only mass that affect is M which is right above m right ?
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LowlyPion
#2
Mar22-09, 03:11 PM
HW Helper
P: 5,341
Quote Quote by nns91 View Post
1. I chose e but I am wrong. What would be the appropriate way to find out the answer of this question ??

2. I said that the distance r^2 is really big so F is negligible at this height. I am wrong again. How should I explain this ??

3. So I did this problem wrong. Is it true that 4 of M are canceled because they are opposite ? So the only mass that affect is M which is right above m right ?
In 1. you have the equation for the period... what happens when you change the mass of the satellite?

In 2. ... what forces are acting on the astronaut?

In 3. ... no. Draw the x,y components of gravitational force.
nns91
#3
Mar22-09, 03:18 PM
P: 301
1. If I double the mass then T^2 will be reduced by a factor of 2. Thus, T will be reduced by square root of 2. ??

2. Gravitational force ??

Delphi51
#4
Mar22-09, 03:21 PM
HW Helper
P: 3,394
Gravity, Kepler's law

All orbit problems should begin with Fc = Fg (the centripetal force is provided by the gravitational force). Putting in the details,
mv^2/R = GmM/R^2 where m is the mass of the satellite, M of the Earth.
or R*v^2 = GM
This is a Kepler's Law for circular orbits. You can use it to answer question 1.

For #2, if the force of gravity was negligible the satellite would move in a straight line rather than in circular motion. Being in a circular orbit means that the force of gravity is exactly right to provide the centripetal force necessary to remain in circular motion. The force accelerates the satellite (and the people inside) toward the center of the Earth with a = v^2/R. This is exactly the same thing that happens when you jump out of an airplane with no parachute (and no significant air resistance). Except that there is a tangential velocity that makes the satellite avoid hitting the Earth as it falls.

Is it true that 4 of M are canceled because they are opposite ? So the only mass that affect is M which is right above m right ?
No. That would apply to 2 of the M's if they are at the extremities of the semicircular arc. The others all have a component that doesn't cancel out. You have to do some serious work on this, using F = GmM/R^2 for each M. Don't forget the angles as you work out the horizontal and vertical components for each F.
Gnosis
#5
Mar22-09, 05:23 PM
P: 143
#2. The spacecraft’s orbital velocity and rate of gravitational attraction per its altitude are such that it allows continuous free-fall precisely per the curvature of the Earth hence altitude is neither gained nor lost. In essence, equilibrium has been established between Earth’s rate of gravitational attraction and per the spacecraft’s orbital velocity allowing endless free-fall with virtually no detectable forces experienced by the spacecraft’s occupants.

This differs considerably from the centripetal force exerted upon an object being rotated on Earth. At the center of a rotating object (such as when a person rotates a ball per a length of string), the person does not produce an attraction toward the center axis as the Earth does to the spacecraft therefore, the only manner in which the rotating ball can be altered from its otherwise tangential direction is per the centripetal force constantly applied via the string as it rotates, which of course, can be felt and measured. So long as the ball on the string maintains a constant velocity per a circular path, it will experience a continuous change in its tangential direction, which produces a continuous acceleration on the rotating ball. The only time the ball does not feel this constant acceleration is if the string is released or the string breaks, in which case, the ball simply travels straight per its tangential direction until it hits the ground. The ball being rotated at a constant velocity via the string is incapable of experiencing equilibrium (no sense of force or acceleration from any direction) while being rotated at a constant velocity, which is totally unlike that of the spacecraft and its occupants per its constant orbital velocity.

So, an object being rotated on the Earth at a constant velocity is still quite different than an object orbiting the planet at a constant velocity hence, the end results of these two concepts are as different as the concepts themselves.
nns91
#6
Mar22-09, 06:09 PM
P: 301
For number 1, is it does not change since velocity is independent of mass of satellite ??
For number 2, then why does the astronaut feel weightless ?
Delphi51
#7
Mar22-09, 07:30 PM
HW Helper
P: 3,394
Yes, you have #1.
For #2 I can only repeat that he is accelerating just like he would after jumping out of an airplane without a parachute.
LowlyPion
#8
Mar22-09, 09:04 PM
HW Helper
P: 5,341
Quote Quote by nns91 View Post
For number 2, then why does the astronaut feel weightless ?
From F = m*a if he is accelerating then if there is something against which the acceleration resists, you know like the floor, if you put a scale there then it will measure a force of something on top, as in the case of the astronaut - his/her weight.

But the question says the astronaut is weightless. So is it the case that gravity is not acting, or is it that there is another force acting such that gravity is counter balanced and the net effect is 0 acceleration (in the frame of reference of the living module)?
nns91
#9
Mar22-09, 09:10 PM
P: 301
Is that because centripetal force and gravitational force are balance so the man is weightless ?
LowlyPion
#10
Mar22-09, 09:19 PM
HW Helper
P: 5,341
Quote Quote by nns91 View Post
Is that because centripetal force and gravitational force are balance so the man is weightless ?
Yes, but don't forget to answer the rest of the question. You can calculate gravitational acceleration is at that height for instance.
Is the force of gravity exerted by the earth on them negligible at this height ?


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