Orbit and Gravity Questions: |
-3 | -2
| -1 |
0 | 1 | 2 | 3
|
4 | 5 | 6 | 7
| 8 | 9 | 10 | 11
| 12 | Go up
- by Chris Murray, 2001 (Problems 5, 6 and 7 are
prefaced with general orbit solving strategy)
-3. Supposing the round up went in a vertical circle, and you measured 3 "g" s at the top, and 5 "g"s at the bottom. How many "g" s is the ride accelerating at, and what is the period of the ride if its radius is 3.6 m
Well, the ride must be pulling 4 "g" s all by itself, at the top one less so we feel 3 "g" s, and at the bottom, one more than 4 "g" s so we feel 5.
4 "g" s is:
ac = 4(9.8 m/s/s) = 39.2 m/s/s
And finally we use the period formula for centripetal acceleration:
ac = 4p2r/T2: ac = 39.2 m/s/s, r = 3.6 m, T = ??, so 39.2 m/s/s = 4p2(3.6 m)/T2
T = 1.9041 = 1.9 s
(Table of contents)
-2. A ride has a radius of 4.5 m, a period of revolution of 2.15 s, and moves in a vertical circle. What is the centripetal acceleration at the edge of the ride, and what "g" force do the riders feel at the top, and at the bottom?
Use ac = 4p2r/T2: ac =??, r = 4.5 m, T = 2.15 s, so ac = 4p2(4.5 m)/(2.15 s)2
ac = 38.4 m/s/s
(38.4 m/s/s)/(9.8 m/s/s) = 3.9 "g"s
At the top you would feel one "g" less or 2.9 "g"s
At the bottom you would feel one "g" more or 4.9 "g"s
(Table of contents)
-1. An amusement park ride has a radius of 4.5 m and is going 11.7 m/s at the top of its vertical circle. What is the centripetal acceleration at the top, and what g-force do the riders feel at the top? What is the minimum speed the ride could go at the top for people to not fall out?
First, let's find the centripetal acceleration:
a = v2/r
a = (11.7 m/s)2/(4.5 m) = 30.42 m/s/s = 30. m/s/s toward the center of the circle.
Now, at the top, the Centripetal acceleration is downward, as the center of the circle lies below the top.
At the top, you feel one less g force than the ride pulls, as earth's gravitational field "cancels" one of the g's.
So the ride pulls (30.42 m/s/s)/(9.80 m/s/s/) = 3.10 "g"s, so you would feel 2.1 "g"s
For the riders to not fall out, you need to pull at least one "g" or have a centripetal acceleration at the top equal to 9.80 m/s/s, so
a = v2/r
(9.80 m/s/s) = v2/(4.5 m), v = 6.6 m/s
(Table of contents)
0. The loop of a roller coaster is 3.8 m. You read 1.7 "g" s at the top of the loop. What is your centripetal acceleration at the top? What is your velocity at the top?
Well, if you are feeling 1.7 "g"s at the top, the ride is pulling one more, or 2.7 "g"s (Because gravity cancels out one g force)
So the acceleration in m/s/s (2.7 "g")(9.80 m/s/s) = 26.46 m/s/s = 26 m/s/s
Your velocity would be:
a = v2/r
(26.46 m/s/s) = v2/(3.8 m), v = 10. m/s
(Table of contents)
1. What is the force of gravity between the two 4.2 Kg bowling balls whose centers are 1.2 m distant?
Using the universal law of gravitation:
F = G(m1m2)/(r2), G = 6.67 x 10-11 Nm2/kg2
F = (6.67 x 10-11 Nm2/kg2)(4.2 kg)(4.2 kg)/(1.2 m)2 = 8.2 x 10-10 N
(Table of contents)
2. What is the force of gravity between the earth and the moon? What acceleration does the moon undergo? (use F=ma) What is its period of motion? (Use a centripetal acceleration equation)
Using the universal law of gravitation:
F = G(m1m2)/(r2), G = 6.67 x 10-11 Nm2/kg2
F = (6.67 x 10-11 Nm2/kg2)(5.98 x 1024 kg)(7.36 x 1022 kg)/(3.80 x 108 m)2 = 2.03 x 1020 N
I can calculate the acceleration of the moon using F = ma
<2.03 x 1022 N> = (7.36 x 1022 kg)a, a = 2.76 x 10-3 m/s/s
And finally, this is the centripetal acceleration of the moon as it moves around the earth, so we can use the formula for centripetal acceleration to calculate the period of its motion:
a = (4π2r)/T2
T2 = (4π2r)/a
T = Ö(4π2r/a) = 2.33 x 106 sec
or ( 2.33 x 106 sec)/(24 hr/day)/(3600 sec/hr) = 27.0 days
(Table of contents)
3. What is the force of gravity on a 10.0 Kg object twice earth's radius above the earth?
The trick here is that you need to use the center to center distance. Since the object is twice earth's radius above earth, it is three earth radii from the center of the earth. (You are most likely one earth radius from the center of the earth right now as you read this). So the center to center distance is three times the earth's radius:
r = 3(6.38x106 m) = 19.14x106 m (The radius of the earth is 6.38x106 m)
And now we can use universal gravitation to find the force of gravity:
F = G(m1m2)/(r2), G = 6.67 x 10-11 Nm2/kg2
F = (6.67 x 10-11 Nm2/kg2)(5.98 x 1024 kg)(10 kg)/(19.14x106 m)2 = 10.9 N
Another trick would be to realize that the force of gravity is 98 N (10 kg times 9.8 N/kg) on the object at one earth radius, if we triple the distance r we are away from the center of the earth (by going to two radii above the earth) you would decrease the force of gravity by an inverse factor of 1/32 = 1/9 the gravity, and 98 N/9 = 10.9 N
(Table of contents)
4. What distance from the center of the moon is the attraction between a 500. kg object and the moon itself equal to 2.5 N
The moon has a mass of 7.36 x 1022 kg, the other mass would be a 500. kg object, so we would use universal gravitation:
F = G(m1m2)/(r2), G = 6.67 x 10-11 Nm2/kg2
Solve for r:
r2 = G(m1m2)/F
r = Ö{G(m1m2)/F} = Ö{(6.67 x 10-11 Nm2/kg2)(7.36 x 1022 kg)(500. kg)/(2.5 N)} = 3.1 x 107 m
(Table of contents)
5. A flea is in orbit 2.1 m from the center of a .545 kg baseball. What is the period of orbit? What is the orbital velocity?
Orbit problems in general:
When a small object orbits an object with much greater mass, the gravitational attraction between the two objects causes both objects to accelerate toward each other, but the smaller accelerates at a much greater rate because of its small mass. In approximation, we will say that the larger mass remains stationary, and the smaller moves in a circle around the larger mass (in the case of a perfect circular orbit)
To make a long story short, for the satellite, Newton's second law has on one side, the force of gravity:
F = G(m1m2)/(r2), G = 6.67 x 10-11 Nm2/kg2
Causing the satellite to undergo a centripetal acceleration: (if it is in a perfect circular orbit) So F = ma looks like:
F = ma
G(m1m2)/(r2) = ma - where "a" is your favorite centripetal acceleration. If we substitute for the centripetal acceleration it would look like:
G(m1m2)/(r2) = m1v2/r
or
G(m1m2)/(r2) = m1(4π2r)/T2
(For both of these, the satellite is m1 because it is the one centripetally accelerating) So now, which one to use? Well it depends on what you know or want to know. The major difference is that one has the period T in it, and the other the velocity v.
This problem:
Now, for the flea orbiting the baseball we know the other mass is the baseball (.545 kg) and the flea's mass is unknown (m1). Since we want to know both the velocity and the period of motion, let's solve for the velocity, since that is mathematically easier, and it seems possible to start with that.
G(m1m2)/(r2) = m1v2/r
Now, I could just put in numbers, but as these are in scientific notation, and sometimes rather large, I would rather do math with symbols, so let's solve for v:
G(m1m2)/(r2) = m1v2/r - (divided both sides by m1) - the flea's mass cancels - that is why we don't need to know it)
G(m2)/(r2) = v2/r - (multiply both sides by r)
G(m2)/(r) = v2
Finally,
v = Ö(Gm2/r) = Ö{(6.67 x 10-11 Nm2/kg2)(.545 kg)/(2.1 m)} = 4.16x10-6 m/s = 4.2x10-6 m/s
Now to find the period, we can either solve the condition of orbit with the period in it:
G(m1m2)/(r2) = m1(4π2r)/T2
Which looks like no fun, or we can be clever.
Velocity, which we already know, is s/t (v = s/t). In a circular orbit, T is the period, or time to complete one revolution. In one revolution, the distance traveled is 2πr, and the time t is T - the period, so you have another relationship which is very helpful:
v = 2πr/T - which we can use in this case to find the Period, since we know v and r:
v = 2πr/T - so
T = 2πr/v = 2π(2.1 m)/(4.16x10-6 m/s) = 3.17x106 s
T = (3.17x106 s)/(3600 s/hr) = 881 hrs = 880 hrs
(Table of contents)
6. What velocity do you need to orbit 8.20 x 106 m from the center of the earth?
Orbit problems in general:
When a small object orbits an object with much greater mass, the gravitational attraction between the two objects causes both objects to accelerate toward each other, but the smaller accelerates at a much greater rate because of its small mass. In approximation, we will say that the larger mass remains stationary, and the smaller moves in a circle around the larger mass (in the case of a perfect circular orbit)
To make a long story short, for the satellite, Newton's second law has on one side, the force of gravity:
F = G(m1m2)/(r2), G = 6.67 x 10-11 Nm2/kg2
Causing the satellite to undergo a centripetal acceleration: (if it is in a perfect circular orbit) So F = ma looks like:
F = ma
G(m1m2)/(r2) = ma - where "a" is your favorite centripetal acceleration. If we substitute for the centripetal acceleration it would look like:
G(m1m2)/(r2) = m1v2/r
or
G(m1m2)/(r2) = m1(4π2r)/T2
(For both of these, the satellite is m1 because it is the one centripetally accelerating) So now, which one to use? Well it depends on what you know or want to know. The major difference is that one has the period T in it, and the other the velocity v.
This problem:
Now, for the you orbiting the earth we know the other mass (m2)is the earth's mass (5.98 x 1024 kg) and your mass is unknown (m1). Since we want to know The velocity of orbit, I choose this orbital condition:
G(m1m2)/(r2) = m1v2/r
Now, I could just put in numbers, but as these are in scientific notation, and rather large, I would rather do math with symbols, so let's solve for v:
G(m1m2)/(r2) = m1v2/r - (divided both sides by m1) - Your mass cancels - that is why we don't need to know it)
G(m2)/(r2) = v2/r - (multiply both sides by r)
G(m2)/(r) = v2
Finally,
v = Ö(Gm2/r) = Ö{( 6.67 x 10-11 Nm2/kg2)(5.98 x 1024 kg)/(8.2 x 106 m)} = 6.9744x103 m/s = 6970 m/s
(Table of contents)
7. What velocity do you need to orbit 5.20 x 105 m from the surface of the moon? What would be your period of orbit?
Orbit problems in general:
When a small object orbits an object with much greater mass, the gravitational attraction between the two objects causes both objects to accelerate toward each other, but the smaller accelerates at a much greater rate because of its small mass. In approximation, we will say that the larger mass remains stationary, and the smaller moves in a circle around the larger mass (in the case of a perfect circular orbit)
To make a long story short, for the satellite, Newton's second law has on one side, the force of gravity:
F = G(m1m2)/(r2), G = 6.67 x 10-11 Nm2/kg2
Causing the satellite to undergo a centripetal acceleration: (if it is in a perfect circular orbit) So F = ma looks like:
F = ma
G(m1m2)/(r2) = ma - where "a" is your favorite centripetal acceleration. If we substitute for the centripetal acceleration it would look like:
G(m1m2)/(r2) = m1v2/r
or
G(m1m2)/(r2) = m1(4π2r)/T2
(For both of these, the satellite is m1 because it is the one centripetally accelerating) So now, which one to use? Well it depends on what you know or want to know. The major difference is that one has the period T in it, and the other the velocity v.
This problem:
Now, for you orbiting the moon we know the other mass (m2) is the moon's mass (7.36 x 1022 kg) and your mass is unknown (m1). Since we want to know both the velocity and the period of motion, let's solve for the velocity, since that is mathematically easier, and it seems possible to start with that.
G(m1m2)/(r2) = m1v2/r
Now, I could just put in numbers, but as these are in scientific notation, and sometimes rather large, I would rather do math with symbols, so let's solve for v:
G(m1m2)/(r2) = m1v2/r - (divided both sides by m1) - the satellite mass cancels - that is why we don't need to know it)
G(m2)/(r2) = v2/r - (multiply both sides by r)
G(m2)/(r) = v2
Finally,
v = Ö(Gm2/r).
There is one last tricky thing about this problem, and that is we are not given the orbital radius which we need to know to solve the problem. We are given the elevation above the moon's surface. To get the distance to the center of the moon, which is what we want for a circular orbit, we need to add the radius of the moon (1738 km - 1,738,000 m). so
r = 5.20 x 105 + 1,738,000 m = 2.258 x 106 m, so we are ready to plug and chug:
v = Ö(Gm2/r) = Ö{( 6.67 x 10-11 Nm2/kg2)(7.36 x 1022 kg)/(2.258 x 106 m)} = 1474 m/s= 1470 m/s
Now to find the period, we can either solve the condition of orbit with the period in it:
G(m1m2)/(r2) = m1(4π2r)/T2
Which looks like no fun, or we can be clever.
Velocity, which we already know, is s/t (v = s/t). In a circular orbit, T is the period, or time to complete one revolution. In one revolution, the distance traveled is 2πr, and the time t is T - the period, so you have another relationship which is very helpful:
v = 2πr/T - which we can use in this case to find the Period, since we know v and r:
v = 2πr/T - so
T = 2πr/v = 2π(2.258 x 106 m)/(1474 m/s) = 9621 s = 9620 s
(Table of contents)
8. Fred the alien orbits the planet Zirkon, completing an orbit with a radius of 23.4 x 106 m every 3.17 earth days. What is his velocity, and what is the mass of the planet he is orbiting?
The orbital velocity is simply the circumference divided by the period
The period = (3.17 days)(24 hr/day)(3600 sec/hr) = 273888 s
The velocity is then:
v = 2πr/T = 2π(23.4 x 106 m)/(273888 s) = 536.8126 m/s = 537 m/s
Now we want the planet mass - So I choose this condition of orbit (I could have chosen either one):
G(m1m2)/(r2) = m1v2/rAnd we want to solve for the mass of the planet which is m2.
Gm2/r2 = v2/r - cancel m1
Gm2/r = v2 - cancel an r
and finally,
m2 = v2r/G = (536.8126 m/s)2(23.4 x 106 m)/(6.67 x 10-11 Nm2/kg2) = 1.01 x 1023 kg
(Table of contents)
9. What is the radius of a geosynchronous orbit? (Around earth, T = 23 hrs, 56 min, 4 sec - but convert it to seconds) How many earth radii is this?
Well, um, yeah, well actually, the given period in seconds is (23 hr)(3600 s/hr) + (56 min)(60 s/min) + (4 s) = 86164 s which is the length of a sidereal day. (It is not 24 hours because that is from high noon to high noon, but in the space of a day, the earth has moved around the sun, so because of the direction the earth rotates about its own axis, and revolves around the sun, it completes one rotation about 1/365th of a day sooner than it takes for the sun to be in the same position again)
Because we have only the mass of the planet, and the period, we are forced to use the second condition of orbit:
G(m1m2)/(r2) = m1(4π2r)/T2
Which we need to solve for r: (There is no other way into this - so here goes:)
G(m1m2)/(r2) = m1(4π2r)/T2
Step 1 - cross multiply:
G(m1m2)T2 = m1(4π2r)(r2)
Step 2 - combine r, and cancel m1:
G(m2)T2 = (4π2r3)
Step 3 - divide both sides by 4π2
G(m2)T2/(4π2) = r3
Step 4 - Cube root both sides
G(m2)T2/(4π2) = r3
3Ö{G(m2)T2/(4π2)} = r
so r = 3Ö{G(m2)T2/(4π2)} = 3Ö{( 6.67 x 10-11 Nm2/kg2)(5.98 x 1024 kg)(86164 s)2/(4π2)} = 4.2174 x 107 m = 4.22 x 107 m
Since the radius of the earth is 6.38 x 106 m, this is (4.2174 x 107 m)/(6.38 x 106 m/radius) = 6.61 times the earth's radius. That's way the heck and gone, if you picture it in your mind...
(Table of contents)
10. What is the period of an orbit 149.6x109 m from the center of the sun (in earth days)?
Well, the mass of the sun is 1.99 x 1030 kg, and we want to use the condition of orbit that has the period in it - so I choose
G(m1m2)/(r2) = m1(4π2r)/T2
And we solve for T:
Gm2/r2 = 4π2r/T2 - cancel m1
Gm2T2 = 4π2r3 = cross multiply
and finally
T = Ö{4π2r3/Gm2} = (31556425.51 s)/(3600 s/hr)/(24 hr/day) = 365.24 days.
HMMM - methinks the planet might be earth...
(Table of contents)
11. At what distance from the surface of the earth is the orbital velocity 5000. m/s?
The mass of the earth is 5.98 x 1024 kg. We want to use the orbit condition with velocity, so we choose:
G(m1m2)/(r2) = m1v2/r
Solving for r:
Gm2/r2 = v2/r - cancel m1 - the satellite mass
Gm2/r = v2 - cancel an r
And finally
r = Gm/v2 = (6.67 x 10-11 Nm2/kg2)(5.98 x 1024 kg)/(5000. m/s)2 = 1.595464 x 107 m
But this is from the center of the earth. We want from the surface, so we subtract the radius of the earth of 6.378 x 106 m and get
1.595464 x 107 m - 6.378 x 106 m = 9.58 x 106 m
(Table of contents)
12. A 4090 kg spacecraft is approaching the moon at 6839 m/s, and wants to orbit 510. km above its surface. What is the speed of orbit at this elevation? They aim 510 km to the side of the moon, but for what time do they need to burn their main engine (that generates 60,123 N of thrust) to reduce their speed to the proper orbital speed?
I guess we need to know first what our orbital velocity is, so let's do that first:
Newton's second law has on one side, the force of gravity:
F = G(m1m2)/(r2), G = 6.67 x 10-11 Nm2/kg2
Causing the satellite to undergo a centripetal acceleration: (if it is in a perfect circular orbit) So F = ma looks like:
F = ma
G(m1m2)/(r2) = ma - where a is your favorite centripetal acceleration. If we substitute for the centripetal acceleration it would look like:
G(m1m2)/(r2) = m1v2/r
or
G(m1m2)/(r2) = m1(4π2r)/T2
(For both of these, the satellite is m1 because it is the one centripetally accelerating) So now, which one to use? Well it depends on what you know or want to know. The major difference is that one has the period T in it, and the other the velocity v.
Now, for the you orbiting the moon we know the other mass (m2) is the moon's mass (7.36 x 1022 kg) and your mass is unknown (m1). Since we want to know the velocity we will use:
G(m1m2)/(r2) = m1v2/r
Now, I could just put in numbers, but as these are in scientific notation, and sometimes rather large, I would rather do math with symbols, so let's solve for v:
G(m1m2)/(r2) = m1v2/r - (divided both sides by m1) - the satellite mass cancels - that is why we don't need to know it)
G(m2)/(r2) = v2/r - (multiply both sides by r)
G(m2)/(r) = v2
Finally,
v = Ö(Gm2/r).
There is one last tricky thing about this part, and that is we are not given the orbital radius which we need to know to solve the problem. We are given the elevation above the moon's surface. To get the distance to the center of the moon, which is what we want for a circular orbit, we need to add the radius of the moon (1738 km). so
r = 510 km + 1738 km = 2248 km = 2248 x 103 m, so we are ready to plug and chug:
v = Ö(Gm2/r) = Ö{( 6.67 x 10-11 Nm2/kg2)(7.36 x 1022 kg)/(2248 x 103 m)} = 1477.7592 m/s
Now that we have this, we need to figure out how to get the spaceship down to that speed. The thrusters 60,123 N of thrust will cause the spaceship to decelerate at F = ma, a = F/m = (60,123 N)/(4090 kg) = 14.7 m/s/s of deceleration. (At the moment of the burn, the force of gravity exerted by the planet is exactly at right angles to the velocity, and since it is in the other direction, I did not incorporate it into my expression of F = ma)
Now we are set, we need to solve a cute linear kinematics question:
x = don't care
vi = 6839 m/s
vf = 1477.7592 m/s
a = -14.7 m/s/s
t = ???
use vf = vi + at, t = 364.7 = 365 s
(Table of contents)