Circular Motion and Gravitation

In summary: No. You cannot just set these two equations equal to each other. To solve for the gravitational acceleration at a point outside the Earth where the force to Earth is one tenth of its value at the Earth's surface, you would need to use the following equation: ##Gm1/r^2 = v2/r##. However, because you do not know the value of v2, you cannot use this equation.
  • #1
NotKepler
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Homework Statement


What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration duo to Earth is 1/10 of its value at the Earth's surface?

Homework Equations


F = Gm1m2/r2
F = m1 (v2/r)
Mass of Earth = 5.98x1024
G = 6.67x10-11

The Attempt at a Solution


From using the two equations stated above I got the equation: Gm1/r2 = v2/r
Since I know that v^2/r is equal to 1/10 the Earth's surface gravity (9.8m/s^2) I got Gm1/r2 = 0.98

Then I plugged everything in and got 20,152,479.97 meters as an answer

Is my work correct?

Thank you,

-Tim
 
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  • #2
Timothy Proudkii said:
From using the two equations stated above I got the equation: Gm1/r2 = v2/r
You cannot just take two equations, slam them together, and hope that you get something out. The expression ##F = mv^2/r## is an expression for the centripetal force in a circular orbit. Nowhere in your problem statement is there anything that talks about a circular orbit. Now, you might get a reasonable answer, but only because the correct identification is ##F = 0.1mg## describes the force necessary to have a tenth of the gravitational acceleration at the surface.

Timothy Proudkii said:
Then I plugged everything in and got 20,152,479.97 meters as an answer
I am sorry, but this is way too many significant digits. There is absolutely no way that you can determine this quantity with that kind of accuracy. In particular given the facts that:
  • Newton's gravitational constant is only known to four significant digits.
  • The mass of the Earth is only known to four significant digits.
  • You have used values for these quantities with only three significant digits.
  • The gravitational acceleration at the Earth's surface cannot said to be exactly 9.8 m/s^2. It varies significantly. Its standard value is defined to be 9.80665 m/s2 and if you are supposed to use that you are making a mistake already in the third significant digit when approximating it by 9.8 m/s^2.
You have 10(!) significant digits. In order to achieve this you would need to know all numbers that enter to a precision that would correspond to knowing your own length to a precision of atomic scales.

Also, please take care to always write out the units of variables when you give them a particular value. Otherwise it has no meaning. For example:
Timothy Proudkii said:
Mass of Earth = 5.98x1024
has no meaning. Without further specification, one has no idea that you are using the unit of kg. For example, you could choose to measure masses in units of the Earth mass and then you would find that the Earth mass is ##1~M_\oplus##, not ##5.98\cdot 10^{24}~M_\oplus##.
 
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  • #3
Orodruin said:
You cannot just take two equations, slam them together, and hope that you get something out. The expression ##F = mv^2/r## is an expression for the centripetal force in a circular orbit. Nowhere in your problem statement is there anything that talks about a circular orbit. Now, you might get a reasonable answer, but only because the correct identification is ##F = 0.1mg## describes the force necessary to have a tenth of the gravitational acceleration at the surface.I am sorry, but this is way too many significant digits. There is absolutely no way that you can determine this quantity with that kind of accuracy. In particular given the facts that:
  • Newton's gravitational constant is only known to four significant digits.
  • The mass of the Earth is only known to four significant digits.
  • You have used values for these quantities with only three significant digits.
  • The gravitational acceleration at the Earth's surface cannot said to be exactly 9.8 m/s^2. It varies significantly. Its standard value is defined to be 9.80665 m/s2 and if you are supposed to use that you are making a mistake already in the third significant digit when approximating it by 9.8 m/s^2.
You have 10(!) significant digits. In order to achieve this you would need to know all numbers that enter to a precision that would correspond to knowing your own length to a precision of atomic scales.

Also, please take care to always write out the units of variables when you give them a particular value. Otherwise it has no meaning. For example:

has no meaning. Without further specification, one has no idea that you are using the unit of kg. For example, you could choose to measure masses in units of the Earth mass and then you would find that the Earth mass is ##1~M_\oplus##, not ##5.98\cdot 10^{24}~M_\oplus##.
If I can't slam these two equations together how else could I solve the problem with the information provided. Could I be using the equations : Net F = ma
and F = Gm1m2/r^2 and set those equal to each other? Sort of what I did previously but instead of using the equation for circular motion.
 
  • #4
Timothy Proudkii said:
Could I be using the equations : Net F = ma and F = Gm1m2/r^2 and set those equal to each other?
You could, but there is an easier way.
Just looking at the second equation, how does gravitational acceleration depend on r? If r is doubled, what happens to the acceleration? How do you have to change r to get acceleration to reduce to one tenth?
 
  • #5
It is better to work with the formulas without explicit numbers (you can insert any actual number at the end, if it is necessary).

From [tex]F=m*a[/tex] and [tex]F=\frac{G*M*m}{r^2}[/tex] you obtain that, for any (small) body of mass "m", in a "Earth-and-body-only" system, the acceleration on the body of mass "m" is:

[tex]a = \frac{G*M}{r^2}[/tex]

If "r_1" is Radius_of_Earth, the "r_2" is what you're looking for, then:


<< Mentor Note -- deleted a bit too much explicit equation help>>

and it is almost done, can you go to the end from here?
 
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Related to Circular Motion and Gravitation

1. What is circular motion?

Circular motion is the movement of an object in a circular path around a fixed point. This type of motion involves both a constant speed and a change in direction.

2. What is centripetal force?

Centripetal force is the force that acts on an object moving in a circular path, directed toward the center of the circle. It is responsible for keeping the object moving in a curved path instead of a straight line.

3. What is the difference between centripetal force and centrifugal force?

Centripetal force is the inward force that keeps an object moving in a circular path, while centrifugal force is the outward force that appears to push an object away from the center of the circle. However, centrifugal force is actually just an apparent force and does not truly exist.

4. How is circular motion related to gravity?

Circular motion and gravity are related through the concept of gravitational force. Gravitational force is responsible for keeping objects in orbit around larger bodies, such as planets or moons. This force acts as the centripetal force that keeps objects moving in a circular path.

5. What is the difference between weight and mass?

Weight is a measure of the force of gravity acting on an object, while mass is a measure of the amount of matter in an object. Weight can change depending on the strength of the gravitational force, while mass remains constant.

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