Gravitational Acceleration Help

In summary: GMm(\frac{1}{r_{1}}-\frac{1}{r_{1}+h}) when h<<r1 near Earth's surface.In summary, the conversation was about proving the practical equality between GMm (1/r1 - 1/r2) and mgh, when r2 is almost equal to r1 near the Earth's surface. The conversation involved using mathematical equations and approximations to show this relationship.
  • #1
wikidrox
44
0
I don't understand how to do this question. Please help

Show that GMm (1/r1 - 1/r2) and mgh are, for practical purposes, equal, when r2 is almost equal to r1 near the Earth's surface.

I just don't know how to do it.
 
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  • #2
1.
[tex]g=\frac{GM}{r_{1}^{2}}[/tex], where [tex]r_{1}[/tex] is the radius of the earth.

2. Set [tex]r_{2}=r_{1}+h[/tex], and assume [tex]h<<r_{1}[/tex]

See if these relations might help you..
 
  • #3
wikidrox said:
I don't understand how to do this question. Please help

Show that GMm (1/r1 - 1/r2) and mgh are, for practical purposes, equal, when r2 is almost equal to r1 near the Earth's surface.

I just don't know how to do it.
Okay, if you're trying to prove that:

[tex]GMm(\frac{1}{r_{1}}\ -\ \frac{1}{r_{2}})\ =\ mgh[/tex]

Notice that you can start by cancelling out "m" from both sides. Also, notice that you can rewrite the stuff in brackets like so:

[tex]GM\frac{r_{2}\ -\ r_{1}}{r_{2}r_{1}}\ =\ gh[/tex]

Now, can you think of a relationship between h, r1, and r2? Also, recognize that r1 is the radius of the Earth. Now, can you use the information from the previous sentence to make an assumption so that you can do something with the r1r2 part?
 
  • #4
hgfdkl;

I still can get it totally.

here is where I get to:

GM/r1^2 + r1h = g

I can't get rid of the r1h.
 
  • #5
Let's do this as follows:

1. You are to approximate:
[tex]GMm(\frac{1}{r_{1}}-\frac{1}{r_{1}+h})[/tex]

2. You know the following:
[tex]\frac{GM}{r_{1}^{2}}=g,h<<r_{1}[/tex]

3. Bring the difference in 1. together like this:
[tex]GMm(\frac{1}{r_{1}}-\frac{1}{r_{1}+h})=GMm(\frac{r_{1}+h-r_{1}}{r_{1}(r_{1}+h)})=[/tex]
[tex]m\frac{GM}{r_{1}^{2}}\frac{h}{1+\frac{h}{r_{1}}}=\frac{mgh}{1+\frac{h}{r_{1}}}[/tex]

4.
Now, by 2:
[tex]h<<r_{1}\to\frac{h}{r_{1}}<<1[/tex]

The fraction in the denominator is seen to be much less than 1, and therefore we have the approximate equality:
[tex]\frac{mgh}{1+\frac{h}{r_{1}}}\approx{mgh}[/tex]
 
  • #6
Then there's always a series expansion :biggrin:
 

1. What is gravitational acceleration?

Gravitational acceleration is the acceleration experienced by an object due to the force of gravity. It is a measure of how quickly an object falls towards the center of a massive body, such as the Earth.

2. How is gravitational acceleration calculated?

Gravitational acceleration can be calculated using the equation a = GM/r^2, where G is the gravitational constant (6.67 x 10^-11 Nm^2/kg^2), M is the mass of the larger object, and r is the distance between the two objects.

3. Does gravitational acceleration vary on different planets?

Yes, gravitational acceleration can vary on different planets depending on their mass and radius. For example, the gravitational acceleration on Earth is 9.8 m/s^2, while on the moon it is only 1.6 m/s^2.

4. What is the difference between gravitational acceleration and free fall acceleration?

Gravitational acceleration is the acceleration due to the force of gravity, while free fall acceleration is the acceleration an object experiences when falling freely in a vacuum. They are equal in magnitude, but free fall acceleration does not take into account air resistance.

5. How does gravitational acceleration affect the motion of objects?

Gravitational acceleration affects the motion of objects by causing them to accelerate towards the center of a massive body. This acceleration is the same for all objects, regardless of their mass, in a vacuum. However, in the presence of air resistance, the acceleration may vary depending on the object's shape and size.

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