What is the gravitational force if the distance is tripled

[ameriland_06]

I am currently in a High School Physics class... i am completely lost with what we are doing, if anyone is really good at physics and has msn messenger please add me at ameriland_06@Hotmail.com If you don't have messenger if you could just e-mail me...

here are some of the questions i have...
1) The gravitational force between two masses is 36 N. What is the gravitational force if the distance is tripled

2) A science fiction writer invents a planet that has twice the radius and eight times the mass of earth. What is the acceleration due to gravity on this planet.

3) A spaceship travels from the Earth directly toward the sun. At what distance from the center of the Earth do the gravitational forces of the sun and the Earth on the ship exactely cancel?

If you could please help me it would be most appreciated b/c this is a major grade and due tomorrow...

 

[chroot]

1) You know Newton's law of universal gravitation, right?

F = G \frac{m_1 m_2}{r^2}

If r is tripled, the denominator of the fraction increased by a factor of nine, and thus the force decreases by a factor of nine.

2) The acceleration due to gravity can be found this way:

F = ma

a = F / m

F = G \frac{M m}{r^2}

a = \frac{G M}{r^2}

Plug in the numbers.

3) You will have one version of Newton's law of universal gravitation describing the force due to the Sun, and another due to the force due to the Earth. Equate them and solve for the distances.

- Warren

 

[ameriland_06]

in the 2nd to last formula what does the Mm stand for?? and wouldn't that be the formula that i would use on 3)

 

[chroot]

ameriland,

In all cases, Newton's law of universal gravitation is applied to two masses. You can call those masses M and m, or m₁ and m₂, or anything else you like.

In problem 2, you're looking for the gravitational acceleration experienced by some test body (call its mass m) in the gravitational field of the earth, with mass M. As you can see, the mass of the test body ends up cancelling out.

- Warren

 

[ameriland_06]

this all is extremely confusing to me... is there a specific site that might help me better understand it all??

 

[chroot]

Ameriland,

You're on the best site there is.

Please explain to me your thought process, and what you think is confusing, and I will try help you.

- Warren

 

[ameriland_06]

2) A science fiction writer invents a planet that has twice the radius and eight times the mass of earth. What is the acceleration due to gravity on this planet.

ok on this problem i did this, a=GM/R squared ... is that the formula i'd use?

3) A spaceship travels from the Earth directly toward the sun. At what distance from the center of the Earth do the gravitational forces of the sun and the Earth on the ship exactely cancel?

On this problem would i use the formula F=G(Mm/r squared) and if i did would the M be the mass of the sun and m be the mass of the earth? i don't really understand on this one?

also do you have any type of messenger so it would be easier to communicate?

 

[chroot]

2) You are correct. Make sure you understand where that formula comes from though; don't just blindly apply it.

3) Consider a test mass m in between the Sun and Earth. Call the Sun's mass M_s and the Earth's mass M_e. The force on the test mass due to the Sun, when at a distance r_s from the Sun, is

F_s = G \frac{M_s m}{r_s^2}

Similarly, the force due to the Earth is

F_e = G \frac{M_e m}{r_e^2}

Solve for when F_s = F_e:

\frac{M_s}{r_s^2} = \frac{M_e}{r_e^2}

You can use now solve for the ratio between the two distances. You'll find that one distance is a multiple of the other. Knowing that the total distance between the Sun and Earth is 93 million miles, you can determine both distances.

- Warren

 

[JasonRox]

For number 3...

You got a=GM/r.

Remember doing something like this...

3x+5y=4
6x+11y=3

...and you had to find x and y.

Well for number 3, you need to find where a equals for the Sun and Earth.

In other words, you get the formula of Earth and the Sun, and you work them together.

 

[ameriland_06]

ok well... i keep getting a wrong answer?? can anyone please tell me the correct answer? and i have more problems

1) to place a satellite into circular orbit 1500 km above the surface of the earth, what velocity must be given to the satellite?

2) what is the ratio of the gravitation force between the moon and the sun to the gravitational force between the Earth and the moon. Is it more accurate to say that the moon orbits the sun or the earth?

also i'd like to check my answers on a few of these...
1) A body orbits the sun at a distance ten time sthe mean distance of Earth's orbit from the sun
a. find the period of the body in years... i got 3.18 X 10^-7
b. Determine the speed of the body... i got 1.49 X 10^-3

2) A 2240-kg communications satellite is in a circular orbit at a height of 35,800 km above the surface of the earth. What is the gravitational force on the satellite? what graction is this of its weight at the surface of the earth?
i got g=9.44 X 10^-8
and i got the weight at the Earth's surface to be 21952 N
*not sure if i did this one right*

3)A science fiction writer invents a planet that has twice the radius and eight times the mass of the earth. What is the acceleration due to gravity on this planet??
i got 2.497 X 10^-22
*don't think this is right either*

 

[JasonRox]

As much as we love physics and math, we still hate doing assignments. We like doing what we want to do.

Unfortunately we like helping people who put an effort in there work.

Did you put any effort?

If yes, show us some.

By giving me what you got, it won't tell me what you did wrong or what you need help on. I can stand here for hours and come up with an infinite amount of possibilities on how you got that answer. That would be a waste of my time, which I am not willing to sacrifice.

Don't get me wrong, I love helping. In fact, I even tutored for free multiple times.

When I tutor, I actually assign homework and I won't tutor the next session until it is done. It can be all wrong, but don't come back with a blank paper that says "I got 4.565423."

Tutors are no different than teachers. They want to see work.

 

[ameriland_06]

welll i don't know how to type out the formulas and what i plugged in?? and if I've put any effort into it, I've been working on this for the past 6 hours straight and i keep getting stuck...

 

[JasonRox]

Have you tried reading the chapter?

 

[chroot]

1) Centripetal acceleration is always a = v^2 / r. Plug in the radius of the orbit (NOT the altitude) and the acceleration due to gravity at that radius, and solve for the velocity.

2) This is just a straightforward application of Newton's law of universal gravitation.

The other answers are basically the same two equations combined in the same ways. Many of your answers are obviously wrong. If a body is orbiting the Sun, but further away than the Earth is, it will be moving slower than the Earth.

Also, remember to put units on your answers. "1.49 X 10^-3" is meaningless.

Can you please show me how you got the answer "g=9.44 X 10^-8" to question number 2?

- Warren

- Warren

 

[ameriland_06]

yes I've read our chapters and they just don't describe these types of problems... do you have any type of messenger so it would be easier for me to communicate to you what I'm doing on these problems??

 

[ameriland_06]

ok... on 2) here is what i did... g=Gm/d(squared) so g=(6.67X10^-11 X 5.97X10^24)/(3.58X10^7 + 6.38X10^6)*squared* forgot to square the distance in my first try... would .2238 N be a better answer?

 

[chroot]

Question 2 was not asking for the gravitational acceleration of the satellite. It was asking for the gravitational force. Just use the law of universal gravitation:

F = G \frac{M m}{r^2}

where M is the mass of the Earth and m is the mass of the satellite.

- Warren

 

[ameriland_06]

ok... i did that formula and got 501.34 N does that sound right? If so to find the weight at the Earth's surface would i just do Mass times Gravity? mass being 2240 kg and the gravity being 501.34 N??

and when you say this 1) Centripetal acceleration is always a = v^2 / r. Plug in the radius of the orbit (NOT the altitude) and the acceleration due to gravity at that radius, and solve for the velocity.

the orbital radius? would that be the Earth's radius plus the altitude??

 

[chroot]

ok... i did that formula and got 501.34 N does that sound right?

Yes.

If so to find the weight at the Earth's surface would i just do Mass times Gravity? mass being 2240 kg and the gravity being 501.34 N??

At the Earth's surface, gravitational acceleration is 9.8 m/s^2. Weight is mg.

the orbital radius? would that be the Earth's radius plus the altitude??

Yes.

- Warren

 

[ameriland_06]

Thanks for everyone's help... I think i now have a better grasp of what's going on