Total Time for Newton's Law of Gravitation

[Phys_Boi]

Homework Statement

Find the total time, t, that an object takes to reach the surface of the Earth from a distance, D, using the Law of Gravitation: $$F_{g} = \frac{GMm}{x^2}$$

R is radius of Earth
D is distance from surface
R+D is total distance from center of masses

****** One Dimension ******

Homework Equations

F = ma
$$\frac{GMm}{x^2} = ma, a = \frac{GM}{x^2}$$

The Attempt at a Solution

$$\frac{a}{v} = \frac{dv}{dx}$$

$$a dx = v dv \frac{-GM}{x^{2}} dx = v dv$$

$$-GM\int_{R}^{R+D} \frac{1}{x^2} dx = \int_{0}^{v} v dv$$

$$v = \sqrt{2GM(\frac{1}{R+D} - \frac{1}{R})}$$

$$\frac{dx}{dt} = \sqrt{2GM(\frac{1}{R+D} - \frac{1}{R})}$$

$$dt = \frac{1}{\sqrt{2GM(\frac{1}{R+D} - \frac{1}{R})}} dx$$

$$\int_{0}^{t} dt = \int_{R}^{R+D} \frac{1}{\sqrt{2GM(\frac{1}{R+D} - \frac{1}{R})}} dx$$$$t = \frac{D}{\sqrt{2GM(\frac{1}{R+D} - \frac{1}{R})}}$$
Is this the correct solution to the problem?

 

[mfb]

A simple cross-check: for D>0, you take the square root of a negative number. That cannot be right.

An unrelated issue: Your speed does not depend on x (the current position) at all. Does that sound realistic?

 

[Phys_Boi]

A simple cross-check: for D>0, you take the square root of a negative number. That cannot be right.

An unrelated issue: Your speed does not depend on x (the current position) at all. Does that sound realistic?

a = GM/r^2

If you treat Earth at the x value of 0 then r is a function of x..?

r(x) = x?

I do understand D>0 causing a problem, but I don't know where I went wrong.

 

[mfb]

R is the constant radius of Earth.

What is r now? The same as x?

I do understand D>0 causing a problem, but I don't know where I went wrong.

Probably a sign error somewhere. Using conservation of energy, you can directly find v(x) and avoid a few steps.

 

[Phys_Boi]

R is the constant radius of Earth.

What is r now? The same as x?
Probably a sign error somewhere. Using conservation of energy, you can directly find v(x) and avoid a few steps.

At time = 0, r = R + D.

I was going to use conservation of energy, but hesitated due to the fact that you use "g", an approximation.
In the first integral, would you use "R" as the top bound instead of "R + D"?

 

[vela]

Homework Statement

Find the total time, t, that an object takes to reach the surface of the Earth from a distance, D, using the Law of Gravitation: $$F_{g} = \frac{GMm}{x^2}$$

R is radius of Earth
D is distance from surface
R+D is total distance from center of masses

****** One Dimension ******

Homework Equations

F = ma
$$\frac{GMm}{x^2} = ma, a = \frac{GM}{x^2}$$

The Attempt at a Solution

$$\frac{a}{v} = \frac{dv}{dx}$$

$$a dx = v dv$$

$$\frac{-GM}{x^{2}} dx = v dv$$

Why'd you throw the minus sign in? I'm not saying it's wrong, but above you wrote $a = \frac{GM}{x^2}$.

$$-GM\int_{R}^{R+D} \frac{1}{x^2} dx = \int_{0}^{v} v dv$$

The bottom limit of the integral is where you're starting from, and the top limit is where you're ending at. That's the convention you followed on the velocity integral, but you did it backwards on the LHS.

Note that if you say your end point is at $x=R$, the $v$ you find is $v(x=R)$, the speed when the object is at a distance $R$. Is that what you want to find?

 

[Phys_Boi]

Why'd you throw the minus sign in? I'm not saying it's wrong, but above you wrote $a = \frac{GM}{x^2}$.


The bottom limit of the integral is where you're starting from, and the top limit is where you're ending at. That's the convention you followed on the velocity integral, but you did it backwards on the LHS.

Note that if you say your end point is at $x=R$, the $v$ you find is $v(x=R)$, the speed when the object is at a distance $R$. Is that what you want to find?

I added the negative just because the drawing that I drew had the object moving left.

I'm trying to find the total time it takes the object to move from (x = R+D) to (x=R).

I realized the mistake and the new equations are:
$$-GM\int_{R+D}^{R} \frac{1}{x^{2}} dx = \frac{v^{2}}{2}$$

$$v = \sqrt{2GM(\frac{1}{R} - \frac{1}{R+D})} = \frac{\mathrm{dx} }{\mathrm{dt}}$$

$$t = \int_{R+D}^{R} \frac{1}{\sqrt{2GM(\frac{1}{R} - \frac{1}{R+D})}} dx$$

$$t = \frac{D}{\sqrt{2GM(\frac{1}{R} - \frac{1}{R+D})}}$$

 

[vela]

As mfb noted earlier, your expression for $v$ doesn't depend on $x$. It's a constant, so you might notice your last calculation amounts to time = distance/speed (with a sign error).

In this problem, you need to pay a bit more attention to the signs. Which way does the velocity point as the object nears the Earth?

 

[Phys_Boi]

As mfb noted earlier, your expression for $v$ doesn't depend on $x$. It's a constant, so you might notice your last calculation amounts to time = distance/speed (with a sign error).

In this problem, you need to pay a bit more attention to the signs. Which way does the velocity point as the object nears the Earth?

Velocity is a function of x. As the object starts off (with a large x value) the velocity is very small. However, as the object gets closer to Earth, the velocity is very great...

The object is traveling to the left, toward the Earth.

 

[vela]

Velocity is a function of x. As the object starts off (with a large x value) the velocity is very small. However, as the object gets closer to Earth, the velocity is very great...

Right, so your expression for $v$ should be a function of $x$. Yours isn't.

The object is traveling to the left, toward the Earth.

According to your sign convention, did you pick the right sign for $v$ when you solved $v^2=$...?

 

[Phys_Boi]

Right, so your expression for $v$ should be a function of $x$. Yours isn't.According to your sign convention, did you pick the right sign for $v$ when you solved $v^2=$...?

I understand now... However, I don't know how to make it a function of x.

And is this right?
$$\frac{-v^{2}}{2}$$

 

[vela]

I understand now... However, I don't know how to make it a function of x.

Think about what I said in post 6.

And is this right?
$$\frac{-v^{2}}{2}$$

No. If you had the equation, $x^2 = 4$, there are two solutions, right? Same thing here. You have $v^2=\text{something}$. Make sure you pick the right solution.

 

[Phys_Boi]

Think about what I said in post 6.No. If you had the equation, $x^2 = 4$, there are two solutions, right? Same thing here. You have $v^2=\text{something}$. Make sure you pick the right solution.

I'm sorry, I still don't understand

 

[mfb]

If you know that x is negative, and x²=4, what is x?