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A carefully designed experiment can measure the gravitational force between masses of \(1 \mathrm{~kg}\). Given that the density of iron is \(7860 \mathrm{~kg} / \mathrm{m}^{3}\) what is the gravitational force between two 1.00 -kg iron spheres that are touching?

Short Answer

Expert verified
Answer: The gravitational force between the two 1 kg iron spheres touching each other can be calculated using the following formula: \(F = (6.674 *10^{-11})\dfrac{(1)(1)}{\left(2*\sqrt[3]{\dfrac{3(1/7860)}{4π}}\right)^2}\) By plugging in the known values and solving, we find the gravitational force between the two spheres.

Step by step solution

01

Determine the volume of the spheres using mass and density

Given the density and mass of the iron spheres, we can find their volumes using the formula: Volume = Mass / Density For 1 kg iron, we have: Density = 7860 kg/m^3 Mass = 1 kg Volume = (1 kg) / (7860 kg/m^3)
02

Find the radius of the spheres using their volume

To find the radius of the sphere, we need to substitute the volume we found in the previous step into the volume formula of the sphere. The formula for the volume of a sphere is: \(V = \dfrac{4}{3}πr^3\) Now, let's solve for r: \(Vr = \dfrac{4}{3}πr^3\) \(r^3 = \dfrac{3V}{4π}\) \(r = \sqrt[3]{\dfrac{3V}{4π}}\) Now plug in the volume from step 1: \(r = \sqrt[3]{\dfrac{3(1/7860)}{4π}}\)
03

Calculate the gravitational force between the two spheres

Now that we know the radius of the spheres, we can use the formula for the gravitational force between two masses. The gravitational force formula is: \(F = G\dfrac{m_1m_2}{d^2}\) Where, F is the force between two masses, G is the gravitational constant (6.674 × 10^-11 m^3⋅kg^-1⋅s^-2), \(m_1\) and \(m_2\) are the masses of the objects, and d is the distance between the centers of the masses. Since the iron spheres are touching each other, the distance between their centers is twice their radius: \(d = 2r\) Now, let's plug the values into the formula: \(F = (6.674 *10^{-11})\dfrac{(1)(1)}{\left(2*\sqrt[3]{\dfrac{3(1/7860)}{4π}}\right)^2}\) After calculating the result, you will find the gravitational force between the two 1 kg iron spheres touching each other.

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Most popular questions from this chapter

With the usual assumption that the gravitational potential energy goes to zero at infinite distance, the gravitational potential energy due to the Earth at its center is a) positive. c) zero. b) negative. d) undetermined.

Two planets have the same mass, \(M,\) but one of them is much denser than the other. Identical objects of mass \(m\) are placed on the surfaces of the planets. Which object will have the gravitational potential energy of larger magnitude? a) Both objects will have the same gravitational potential energy. b) The object on the surface of the denser planet will have the larger gravitational potential energy. c) The object on the surface of the less dense planet will have the larger gravitational potential energy. d) It is impossible to tell.

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A space shuttle is initially in a circular orbit at a radius of \(r=6.60 \cdot 10^{6} \mathrm{~m}\) from the center of the Earth. A retrorocket is fired forward reducing the total energy of the space shuttle by \(10.0 \%\) (that is, increasing the magnitude of the negative total energy by \(10.0 \%\) ), and the space shuttle moves to a new circular orbit with a radius that is smaller than \(r\). Find the speed of the space shuttle (a) before and (b) after the retrorocket is fired.

A spherical asteroid has a mass of \(1.769 \cdot 10^{20} \mathrm{~kg} .\) The escape speed from its surface is \(273.7 \mathrm{~m} / \mathrm{s}\). What is the radius of the asteroid?

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