On Earth, a person can jump vertically and rise to a height h . What is the radius of the larges

On Earth, a person can jump vertically and rise to a height...

Key Concepts

Kinetic Energy and Jump Height

When a person jumps vertically, their initial kinetic energy is converted into gravitational potential energy at the peak of the jump. Under Earth’s gravity, this relationship is given by (1/2)mv² = mgh, where h is the maximum height reached. This concept helps in comparing the energy available from a jump to the energy needed to overcome another body's gravitational field.

Uniform Density Sphere

A uniform density sphere is a model where the mass is evenly distributed throughout the volume of the sphere. For such a sphere, the total mass can be expressed as M = (4/3)?R³?, where ? is the density and R is the radius. This relation is fundamental when linking the size (radius) of an asteroid or planet to its gravitational characteristics.

Conservation of Energy in Gravitational Fields

This concept involves the conversion of kinetic energy into gravitational potential energy and vice versa. By equating the energy an object has when it is thrown (or jumped) upward to the energy needed to overcome the gravitational pull of a body, one can determine quantities like the escape velocity or the maximum height achievable in a gravitational field.

Escape Velocity

Escape velocity is the minimum speed an object must have to escape from a celestial body's gravitational pull without further propulsion. It is derived from equating the kinetic energy required to the gravitational potential energy of a spherical mass, and is typically given by the formula v? = ?(2GM/R), where G is the gravitational constant, M is the mass of the body, and R is its radius.

Transcript

00:01 When a person jumps vertically on earth, they can rise to a height of h.

00:05 We are given that the density of an asteroid is 3 ,500 kilograms per cubic meter, and we want to find the largest radius of a spherical asteroid in which the person could escape by just jumping straight upwards.

00:22 Here i've drawn a diagram of the earth on the left with a person on the surface jumping to a height h.

00:28 And on the right there's an asteroid with the person on the surface and at the top it represents when the person reaches infinity and let's go ahead and label these positions one two three and four so here we're going to use the conservation of energy so energy at one is equal to energy at two let's go ahead and break those into the energy components the cannot energy at 1 plus the gravitational potential energy at 1 is equal to the kinetic energy at 2 plus the gravitational potential energy at 2.

01:12 So the kinetic energy at 1 will be 1 half mv squared sub 1 minus or plus the gravitational potential energy term which is negative g.

01:29 M of the person, m earth, divided by the radius of the earth, since that's the distance between the person and the center of the earth, this will be equal to the kinetic energy at the top, which is zero because the person has reached the top of his or her trajectory, plus or minus g, m, person, m, earth, divided by the radius of the earth plus the height.

02:02 If we solve for the kinetic energy, we get g m, m, m, e, times 1 over r .e, minus 1 over r .e plus h.

02:22 So now let's go ahead and look at situation 2, where we have energy at 3 is equal to energy at 4.

02:32 Energy at 3, again we have kinetic energy.

02:37 And actually, this kinetic energy that we start with at the surface of the asteroid is the same as when we start at the earth.

02:45 So using that k .e .1 is equal to k .e3, we can go ahead and substitute it in in the second line, plus gravitational potential energy at 3 equals energy of 4 will be at infinity.

03:03 So that's when the person finally reaches rest and is infinitely far from the asteroid.

03:09 So both kinetic and gravitational potential energy is zero.

03:14 So let's go ahead and plug in what we had for kinetic energy 1 in here.

03:24 So that we have g .m .m .e.

03:29 1 over r .e.

03:31 Minus 1 over r .e plus h.

03:36 And on the right hand side, let's go ahead and move u3 to the other side, and the negative -negative cancels so that we have g, m, m, a, this time that's the mass of the asteroid, divided by the radius of the asteroid, since that's the distance between the person and the center of the asteroid when the person is at the surface of the asteroid.

04:00 So we can cancel out a couple terms, specifically the g and the m, and, and, and, you can cancel out.

04:08 And notice over here that we have this density term.

04:15 So let's go ahead and make this a little neater.

04:21 M .e.

04:24 1 over r .e.

04:26 Minus 1 over r .e plus h equals m .a over r .a.

04:35 So if we divide this component, so this component, to this side, we get 1 equals 3 ,500 divided by everything over here.

04:52 So here we can pretend that we are multiplying by 1.

04:56 So let's go ahead and plug in what 1 is over here, which we said was 3 ,500 times 4 pi r a cubed divided by ma.

05:16 And we can cancel some terms here...