Circular Motion and Gravitation

[NotKepler]

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

 

[Orodruin]

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.

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:

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 $5.98\cdot 10^{24}~M_\oplus$.

 

[NotKepler]

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.

 

[haruspex]

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?

 

[mattt]

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 $$F=m*a$$ and $$F=\frac{G*M*m}{r^2}$$ 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:

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

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