Problem Physics I : gravitation

In summary: So the formula F_g=G\frac{M_Em}{R^2} is valid only for a body of mass 1kg? If it's the case then why not say it clearly in the statement of the problem? OK, thanks for your help!In summary, the problem involves calculating the universal gravitation constant given the radius and density of the Earth, as well as the acceleration of gravity on its surface. The formula F_g=\frac{GM_Em}{R_E^2} is used, with F_g representing the gravitational force on a mass m at the Earth's surface. After plugging in the given values and solving for G, the correct answer is obtained. It is important to note that the formula
  • #1
fluidistic
Gold Member
3,953
265

Homework Statement


Suppose that the Earth is a sphere with a radius of [tex]6371 \text{ km}[/tex] and that its uniform density is worth [tex]5517 \text{ kg}/m^3[/tex]. Suppose also that the acceleration of the gravity on its surface is [tex]g=9.80665 m/s^2[/tex]. Calculate the value of the universal gravitation constant.
(The answer should be [tex]G=6.672 \cdot 10^{-11}Nm^2/kg^2[/tex].)


Homework Equations

[tex]F_g=\frac{GM_Em}{R_E^2}[/tex].



The Attempt at a Solution


Using simple very well known formulae, I could determine the mass of the Earth to be about [tex]5.97 \cdot 10^{24}kg[/tex].
From [tex]F_g=\frac{GM_Em}{R_E^2}[/tex] I got that [tex]G=\frac{R_E^2F_g}{M_Em}[/tex]. Now the problem is that I got [tex]G=6.6607246 \cdot 10^{-11}m^3/(kg^2s^2)=6.6607246 \cdot 10^{-11}Nm^2/kg^2[/tex] as I should but I reached this because I supposed that in the formula m=1kg and the body whose mass is 1kg is on the ground of the Earth. Why do I reach the result when I supposed that there is a mass of 1kg on the ground? Because to use the formula you have to have 2 bodies, the Earth and another body. In my case I supposed it was a body with a mass of 1kg and it worked. But if it had a different mass the result would have been totally different. Also, there's no mention of another body (nor even the formula to calculate the universal gravitational constant) in the statement of the problem. I'm certainly missing plenty of things... Could you explain to me what I don't understand? Thanks in advance.
 
Physics news on Phys.org
  • #2
Hi fluidistic,

fluidistic said:

Homework Statement


Suppose that the Earth is a sphere with a radius of [tex]6371 \text{ km}[/tex] and that its uniform density is worth [tex]5517 \text{ kg}/m^3[/tex]. Suppose also that the acceleration of the gravity on its surface is [tex]g=9.80665 m/s^2[/tex]. Calculate the value of the universal gravitation constant.
(The answer should be [tex]G=6.672 \cdot 10^{-11}Nm^2/kg^2[/tex].)


Homework Equations

[tex]F_g=\frac{GM_Em}{R_E^2}[/tex].



The Attempt at a Solution


Using simple very well known formulae, I could determine the mass of the Earth to be about [tex]5.97 \cdot 10^{24}kg[/tex].
From [tex]F_g=\frac{GM_Em}{R_E^2}[/tex] I got that [tex]G=\frac{R_E^2F_g}{M_Em}[/tex]. Now the problem is that I got [tex]G=6.6607246 \cdot 10^{-11}m^3/(kg^2s^2)=6.6607246 \cdot 10^{-11}Nm^2/kg^2[/tex] as I should but I reached this because I supposed that in the formula m=1kg and the body whose mass is 1kg is on the ground of the Earth. Why do I reach the result when I supposed that there is a mass of 1kg on the ground? Because to use the formula you have to have 2 bodies, the Earth and another body. In my case I supposed it was a body with a mass of 1kg and it worked. But if it had a different mass the result would have been totally different. Also, there's no mention of another body (nor even the formula to calculate the universal gravitational constant) in the statement of the problem. I'm certainly missing plenty of things... Could you explain to me what I don't understand? Thanks in advance.

What did you plug in for [itex]F_g[/itex] when you solved for [itex]G[/itex]? If you did that correctly I think you'll see why it doesn't matter what [itex]m[/itex] is.
 
  • #3
What did you plug in for LaTeX Code: F_g when you solved for LaTeX Code: G ? If you did that correctly I think you'll see why it doesn't matter what LaTeX Code: m is.
I plugged [tex]9.80665m/s^2[/tex] for [tex]F_c[/tex]. I know that there is "m" at the denominator but as it is a mass, its unit is not m but kg. So I still don't see why it doesn't matter what m is...
 
  • #4
fluidistic said:
I plugged [tex]9.80665m/s^2[/tex] for [tex]F_c[/tex]. I know that there is "m" at the denominator but as it is a mass, its unit is not m but kg. So I still don't see why it doesn't matter what m is...

The value [tex]9.80665m/s^2[/tex] is an acceleration so it can't be [itex]F_g[/itex]. The force [itex]F_g[/itex] is the gravitational force (weight) that the mass [itex]m[/itex] experiences when it is at a place where the graviational accleration is equal to [itex]g[/itex]. So for a mass [itex]m[/itex] at the surface of the earth, what would [itex]F_g[/itex] be?
 
  • #5
[tex]F_g=mg[/tex] in that case! Thank you so much, I corrected the units and all work perfectly now.
 

FAQ: Problem Physics I : gravitation

What is the force of gravity?

The force of gravity is the attractive force between two objects with mass. It is a fundamental force of nature that is responsible for holding the planets in orbit around the sun and keeping objects on Earth's surface.

How does mass affect gravitational force?

The greater the mass of an object, the greater its gravitational force. This means that larger objects have a stronger gravitational pull than smaller objects. For example, the sun has a much larger mass than Earth, which is why it has a stronger gravitational pull on our planet.

What is the equation for gravitational force?

The equation for gravitational force is F = G * (m1 * m2) / r^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

How does distance affect gravitational force?

The farther apart two objects are, the weaker the gravitational force between them. This is because the force of gravity decreases with distance. If you double the distance between two objects, the force of gravity will decrease by a factor of 4.

What is the difference between mass and weight?

Mass is a measure of the amount of matter in an object, while weight is a measure of the force of gravity on an object. Mass is constant and does not change, while weight can vary depending on the strength of the gravitational pull. On Earth, an object's weight is equivalent to its mass multiplied by the acceleration due to gravity (9.8 m/s^2).

Similar threads

Replies
14
Views
1K
Replies
3
Views
1K
Replies
5
Views
2K
Replies
8
Views
1K
Replies
8
Views
1K
Back
Top