Earth's density and its gravity

[new-tonian]

Homework Statement

Jack is tired of having to use 9.81 m/s^2 all the time to his calculations concerning gravity. He wants to use a round number 10 m/s^2. How much mass would have to be added to the Earth to make this happen? What would the new radius of the Earth be? Assume that the density of the Earth will remain the same. Where could you get the mass needed?

The Attempt at a Solution

All I have right now is the formula p=m/v. I know the volume of the Earth, its density and its mass, but I can't find the relationship between mass/density/volume and gravity.
I have to find the answer guided by the density formula only. I've tried to multiply the mass by the ratio of 9.81 and 10, but somehow the answer doesn't look right, and I won't know if it's right if I don't know how the 9.81 m/s^2 is derived from the density formula.

I'm not looking for the answer, just for hints as to how to approach this problem. Thanks!

 

[hage567]

What equation do you use for gravitational force? If you think about that and Newton's second law, you can see where "g" comes from.

 

[Moetasim]

I understand Jack's frustration with using a specific value for the gravitational acceleration of Earth all the time. However, it is important to note that the value of 9.81 m/s^2 is not arbitrary and is actually derived from the density and mass of Earth.

To understand this, we need to look at the universal law of 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. This can be expressed mathematically as F = G(m1m2)/r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

In the case of Earth, the gravitational force acting on an object on its surface is due to the mass of Earth, which is concentrated at its center. This can be expressed as F = G(m1m2)/r^2, where m1 is the mass of Earth and m2 is the mass of the object on its surface. The distance between the object and Earth's center (r) is equal to the radius of Earth (R).

Now, let's look at the density formula you mentioned, p = m/v. This formula tells us that the mass of an object (m) is equal to its density (p) multiplied by its volume (v). In the case of Earth, we can express its mass as m = pV, where V is the volume of Earth.

Substituting this into our previous equation for gravitational force, we get F = G(pVm2)/R^2. Now, let's consider what happens when we increase the gravitational acceleration from 9.81 m/s^2 to 10 m/s^2. This means that we are increasing the force of gravity acting on an object on Earth's surface. In order to do this, we need to increase the mass of Earth (m1) by a certain amount (Δm).

Therefore, we can express the new gravitational force as F = G((pV+Δm)m2)/R^2. We can then equate this to the new gravitational acceleration we want, 10 m/s^2, and solve for Δm.

10 = G((pV+Δm)m2)/R^2

Δm = (10R^2)/(G