A newly-discovered plant, "Cosmo", has a mass that is 4 times the mass of the Earth. The radius of the Earth is Re. The gravitational field strength at the surface of Cosmo is equal to that at the surface of the Earth if the radius of Cosmo is equal to...

I guess I need some kind of relationship between mass, radius, and gravity. Originally I thought that since Fg=ma and a=v^2/r for circular motion (IS this circular motion???) then Fg=mv^2/r. But then if m is multiplied by 4, so is r. But that's wrong...

You are on the right track with using the equation for gravitational force, but let me clarify the relationships between mass, radius, and gravitational field strength for you.

The force of gravity (Fg) between two objects is given by the equation Fg = G * (m1 * m2) / r^2, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between their centers.

The gravitational field strength (g) at a point in space is defined as the force per unit mass experienced by an object at that point. So, the gravitational field strength can be calculated by dividing the force of gravity (Fg) by the mass of the object (m).

In this case, we want to compare the gravitational field strength at the surface of Cosmo to that at the surface of the Earth. Since both are at their respective surfaces, the mass of the object cancels out. Thus, we can use the equation g = G * (m / r^2) to find the relationship between the radii of the two objects.

Let's denote the radius of the Earth as Re and the radius of Cosmo as Rc. We can set up the equation as follows:

G * (mE / Re^2) = G * (mC / Rc^2)

Since the gravitational field strength is equal at the surfaces of both objects, we have:

(mE / Re^2) = (mC / Rc^2)

Given that the mass of Cosmo is 4 times the mass of the Earth, we can substitute mC = 4mE:

(mE / Re^2) = (4mE / Rc^2)

We can simplify further by canceling out the mass of the Earth:

(1 / Re^2) = (4 / Rc^2)

From this equation, we can find the relationship between the radii of the Earth and Cosmo:

Re^2 = (4 / 1) * Rc^2

Simplifying:

Re = 2 * Rc

Therefore, the radius of Cosmo would be half the radius of the Earth (Re/2) in order for the gravitational field strength at its surface to be equal to that at the surface of the Earth.

To find the relationship between mass, radius, and gravitational field strength, we need to use Newton's law of universal gravitation. This law states that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The formula for the gravitational field strength, g, at the surface of a planet is given by:

g = (G * M) / R^2

where G is the gravitational constant (approximately 6.67 x 10^-11 N m^2 / kg^2), M is the mass of the planet, and R is the radius of the planet.

Given that the mass of Cosmo is 4 times the mass of Earth (M_Cosmo = 4 * M_Earth) and the radius of Earth is Re, we need to find the radius of Cosmo (R_Cosmo) such that the gravitational field strength at the surface of Cosmo (g_Cosmo) is equal to the gravitational field strength at the surface of Earth (g_Earth).

Setting up the equation:

g_Cosmo = (G * M_Cosmo) / R_Cosmo^2
g_Earth = (G * M_Earth) / Re^2

Since the two gravitational field strengths are equal, we can set them equal to each other:

(G * M_Cosmo) / R_Cosmo^2 = (G * M_Earth) / Re^2

Now we can substitute M_Cosmo = 4 * M_Earth and simplify the equation:

(G * (4 * M_Earth)) / R_Cosmo^2 = (G * M_Earth) / Re^2

Cancelling out G and M_Earth:

(4 / R_Cosmo^2) = (1 / Re^2)

Cross-multiplying:

4 * Re^2 = R_Cosmo^2

Now we can solve for the radius of Cosmo (R_Cosmo):

R_Cosmo^2 = 4 * Re^2
R_Cosmo = 2 * Re

Therefore, the radius of Cosmo is 2 times the radius of Earth.

You need Fg = G m M2/r^2

What the problem says is that the force on a sample mass m is the same at the surface.

Fg earth = G m Mearth/Rearth^2
Fg cosmo = G m 4 Mearth/ Rcosmo^2
they are the same for the same little m
so

4 /Rcosmo^2 = 1/Rearth^2

Rcosmo = 2 Rearth