Acceleration Part 2: Obtain Expression for g in Terms of G, Ra & ρ

In summary, to obtain an expression for ga, we can use the equation F = G*m1*m2/r^2 and substitute the volume of a sphere into the equation. This gives us the equation F = G*m2*p*4*PI*r/3, where p is the density of the asteroid. Solving for g, we get g = G*p*4*PI*r/3. Therefore, ga = G*p*4*PI*r/3 in terms of G, Ra, and ρ. To determine Ra, we can use the given value of g and solve for Ra using the equation g = G*m/Ra^2.
  • #1
322
0

Homework Statement



density, ρ of the rock is 4000 kg m−3 and that this density is uniform throughout. You have equipment that enables you to measure the time it takes for a small object to fall from rest through a distance of 2.00 m. Your measurement is 8.0 s.

Obtain an expression for ga in terms of the gravitational constant G, the
radius of the asteroid Ra and the density ρ of the asteroid. Hence determine
Ra.

Homework Equations



F= Gm/ r2

F=mg

volume of sphere= 4/3 * pi * r^3

density= m/v

The Attempt at a Solution



substituting volume of sphere= 4/3 * pi * r^3 into density= m/v gives-

density= 3m/ ( 4* pi * r^3 )

4/3 * pi * r^3 * density = m

F= Gm/ r2

F= (G*4/3 * pi * r^3 * density) / r^2

F=G*4/3 * pi * r * density

F=mg

mg= G*4/3 * pi * r * density

4/3 * pi * r^3 * density = m

4/3 * pi * r^3 * density* g = G*4/3 * pi * r * density

g= Gr^-2

or g= G / r^2


(is this correct)?
 
Physics news on Phys.org
  • #2
F= Gm/ r2 ?

No.
F = G*m1*m2/r^2
 
  • #3
if i use F = G*m1*m2/r^2, i am only given 1 object, 1 mass...

so how would i go about doing this by using

F = G*m1*m2/r^2
 
  • #4
You don't need to know the other mass, but you must include it in your equations. You'll see why.
 
  • #5
imy786 said:

Homework Statement



density, ρ of the rock is 4000 kg m−3 and that this density is uniform throughout. You have equipment that enables you to measure the time it takes for a small object to fall from rest through a distance of 2.00 m. Your measurement is 8.0 s.

Obtain an expression for ga in terms of the gravitational constant G, the
radius of the asteroid Ra and the density ρ of the asteroid. Hence determine
Ra.

Homework Equations



F= Gm/ r2

F=mg

volume of sphere= 4/3 * pi * r^3

density= m/v

The Attempt at a Solution



substituting volume of sphere= 4/3 * pi * r^3 into density= m/v gives-

density= 3m/ ( 4* pi * r^3 )

4/3 * pi * r^3 * density = m

F= Gm/ r2

F= (G*4/3 * pi * r^3 * density) / r^2

F=G*4/3 * pi * r * density

F=mg

mg= G*4/3 * pi * r * density

4/3 * pi * r^3 * density = m

4/3 * pi * r^3 * density* g = G*4/3 * pi * r * density

g= Gr^-2

or g= G / r^2


(is this correct)?

As others pointed out, you are not using the correct equation for the force of gravity!

be very careful, there are two masses involved here, th etotal mass of the "rock and the mass of the falling object. You should really use two different symbols and be careful about using the correct one in each eqaution you are using.

As for your final answer, there is an obvious simple check to make...does it have the dimensions of an acceleration??
 
  • #6
F = G*m1*m2/r^2

substituting volume of sphere= 4/3 * pi * r^3 into density= m/v gives-

density= 3m/ ( 4* pi * r^3 )

4/3 * pi * r^3 * density = m

im stuck here...

shall i definte 4/3 * pi * r^3 * density = m = m1

and leave m2 as m2
 
  • #7
Sure. Just keep them consistent.
 
  • #8
F = G*m1*m2/r^2

V= (4/3)*PI*r^3

p= m /v

p= m1/ (4/3)*PI*r^3

m1= p*4*PI*r^3/ 3

now we substiute this value of m1 to this equation F = G*m1*m2/r^2

F = G*m2*p*4*PI*r/ 3

is final answer...can someone tell me if this correct..to Obtain an expression for ga in terms of the gravitational constant G, the
radius of the asteroid Ra and the density ρ of the asteroid. Hence determine
Ra.
 
  • #9
F = G*m2*p*4*PI*r/ 3

F=mg

mg= G*m2*p*4*PI*r/ 3

g= G*p*4*PI*r/ 3

as final answer...can anyone..confirm if this is correct formula

ga in terms of the gravitational constant G, the
radius of the asteroid Ra and the density ρ of the asteroid.
 

1. What is the significance of obtaining an expression for g in terms of G, Ra, and ρ?

Obtaining an expression for g in terms of these variables allows us to understand and quantify the acceleration due to gravity in different situations, such as on different planets or in different materials. It also allows us to make predictions and calculations related to gravity and its effects.

2. How is the expression for g derived?

The expression for g is derived from Newton's law of universal gravitation, which states that the force of gravity between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. By manipulating this equation and incorporating the variables G, Ra, and ρ, we can obtain an expression for g.

3. What is the role of G in the expression for g?

G is the universal gravitational constant, which is a fundamental constant in physics. It represents the strength of the gravitational force between two objects with a given mass and distance. Its value is approximately 6.67 x 10^-11 N*m^2/kg^2.

4. How does the density (ρ) of a material affect the value of g?

The density of a material affects the value of g because it is a measure of how much mass is present in a given volume. The more mass there is in a material, the more gravitational force it exerts, resulting in a higher value of g. Therefore, the denser the material, the higher the value of g will be.

5. Can the expression for g be used to calculate the acceleration due to gravity on other planets?

Yes, the expression for g can be used to calculate the acceleration due to gravity on other planets, as long as we have the necessary values for G, Ra, and ρ for that specific planet. This allows us to compare the strength of gravity on different planets and understand how it varies based on factors like mass and density.

Suggested for: Acceleration Part 2: Obtain Expression for g in Terms of G, Ra & ρ

Back
Top