Gravitational Field Question

In summary: So what do you get when you solve for r? That will give you the radius of the Earth plus the altitude of the meteor. To get the altitude only you need to subtract the radius of the Earth from that.For the second question, you don't need to solve anything. You are given F= ma, so all you need to do is to find a for the 30 kg meteor and multiply it by 30.
  • #1
EE123
5
0

Homework Statement


A 12 kg meteor experiences an acceleration of 7.2 m/s^2, when falling towards the earth.

a.) How high above the Earth's surface is the meteor?

b.) What force will a 30 kg meteor experience at the same altitude?


Homework Equations



Fg = ma , Fg =Gm1m2 / r^2




The Attempt at a Solution



a = Fg / m, thus:

Fg / m = Gm1m2 / r^2

then:

r = √ (Fg / Gm1m2(m))

I got this far, somewhat understanding what I was doing the equation manupulation. Please help me! I searched this question up, but that thread was not helpful.

I don't know how to solve the question. The altitude is the radius from the earth? and since a = Fg / m, :S I equated this with the Fg = Gm1m2 / r^2, :|.

Please help!
 
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  • #2
EE123 said:

Homework Statement


A 12 kg meteor experiences an acceleration of 7.2 m/s^2, when falling towards the earth.

a.) How high above the Earth's surface is the meteor?

b.) What force will a 30 kg meteor experience at the same altitude?


Homework Equations



Fg = ma , Fg =Gm1m2 / r^2




The Attempt at a Solution



a = Fg / m, thus:

Fg / m = Gm1m2 / r^2

then:

r = √ (Fg / Gm1m2(m))

I got this far, somewhat understanding what I was doing the equation manupulation. Please help me! I searched this question up, but that thread was not helpful.
Do you understand what "Gm1m2/r^2" means? What is G? What are m1 and m2?
A really major error is setting "a= Fg/m= Gm1m2/r^2". "Gm1m2/r^2" is the force due to gravity, not the acceleration. What you want is Fg= ma= Gm1m2/r^2.

Now, again, what are m1 and m2?

I don't know how to solve the question. The altitude is the radius from the earth? and since a = Fg / m, :S I equated this with the Fg = Gm1m2 / r^2, :|.
That wolyld be correct but you didn't! You equated it to a, not Fg.

Please help!
 
  • #3
HallsofIvy said:
Do you understand what "Gm1m2/r^2" means? What is G? What are m1 and m2?
A really major error is setting "a= Fg/m= Gm1m2/r^2". "Gm1m2/r^2" is the force due to gravity, not the acceleration. What you want is Fg= ma= Gm1m2/r^2.

Now, again, what are m1 and m2?


That wolyld be correct but you didn't! You equated it to a, not Fg.


Yes I do understand what the equation: Fg = Gm1m2 / r^2 this is what my notes say (taking an online physics course, they provide notes)

G = gravitational field constant, m1 and m2 are the two masses, Fg = the force of gravity acting on each object,

r = the distance between the centres of the two masses.

this is the Newtons law of universal gravitation in the above equation form.

So ma = Gm1m2 / r^2 ??

which means r = √ (ma / Gm1m2)?

m1 would be the Earth's mass and m2 would be the meteor mass? but the mass, m2, would cancel out right?
 
  • #4
EE123 said:
Yes I do understand what the equation: Fg = Gm1m2 / r^2 this is what my notes say (taking an online physics course, they provide notes)

G = gravitational field constant, m1 and m2 are the two masses, Fg = the force of gravity acting on each object,

r = the distance between the centres of the two masses.

this is the Newtons law of universal gravitation in the above equation form.

So ma = Gm1m2 / r^2 ??

which means r = √ (ma / Gm1m2)?

m1 would be the Earth's mass and m2 would be the meteor mass? but the mass, m2, would cancel out right?
It looks as if you have three different masses, m1, m2 and m. There are only two objects, so there should only be two different masses.
 
  • #5
EE123 said:
ma = Gm1m2 / r^2 ??

which means r = √ (ma / Gm1m2)?

m1 would be the Earth's mass and m2 would be the meteor mass? but the mass, m2, would cancel out right?

The top equation is right, but you solved for r wrong.
And you are right about m2, it does cancel.
 

1. What is a gravitational field?

A gravitational field is a region in space where a mass experiences a force due to the presence of another mass. The strength of the gravitational field is determined by the masses of the objects and the distance between them.

2. How is the strength of a gravitational field measured?

The strength of a gravitational field is measured by its gravitational acceleration, which is the rate at which an object falls towards the source of the field. It is usually measured in meters per second squared (m/s²).

3. How does the mass of an object affect the gravitational field?

The greater the mass of an object, the stronger its gravitational field will be. This means that objects with larger masses will exert a stronger force on other objects compared to objects with smaller masses.

4. What is the relationship between distance and gravitational field?

The strength of a gravitational field decreases as the distance between two objects increases. This means that the farther away two objects are from each other, the weaker the gravitational force between them will be.

5. How does the concept of gravitational field relate to Newton's Law of Universal Gravitation?

The concept of gravitational field is based on Newton's Law of Universal Gravitation, which states that every object in the universe attracts every other object with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

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