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(II) Every few hundred years most of the planets line up onthe same side of the Sun. Calculate the total force on the Earthdue to Venus, Jupiter, and Saturn, assuming all four planetsare in a line, Fig. $24 .$ The masses are $M_{\mathrm{V}}=0.815 M_{\mathrm{E}}$$M_{\mathrm{J}}=318 M_{\mathrm{E}}, M_{\mathrm{Sat}}=95.1 M_{\mathrm{E}},$ and the mean distances of the four planets from the Sun are $108,150,778,$ and1430 million $\mathrm{km} .$ What fraction of the Sun's force on theEarth is this?

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27 millionths fraction.

Physics 101 Mechanics

Chapter 6

Gravitation and Newton's Synthesis

Physics Basics

Newton's Laws of Motion

Applying Newton's Laws

Gravitation

Rutgers, The State University of New Jersey

University of Washington

University of Winnipeg

Lectures

03:43

In physics, dynamics is the branch of physics concerned with the study of forces and their effect on matter, commonly in the context of motion. In everyday usage, "dynamics" usually refers to a set of laws that describe the motion of bodies under the action of a system of forces. The motion of a body is described by its position and its velocity as the time value varies. The science of dynamics can be subdivided into, Dynamics of a rigid body, which deals with the motion of a rigid body in the frame of reference where it is considered to be a rigid body. Dynamics of a continuum, which deals with the motion of a continuous system, in the frame of reference where the system is considered to be a continuum.

03:55

In physics, orbital motion is the motion of an object around another object, which is often a star or planet. Orbital motion is affected by the gravity of the central object, as well as by the resistance of deep space (which is negligible at the distances of most orbits in the Solar System).

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(II) Every few hundred yea…

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Every few hundred years mo…

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The mass of the Sun is $2 …

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Find the gravitational for…

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so here. In order to calculate the net force than that, don't gravitational force of the planets on the earth. We need to find the distance between these planets and the earth. So we can say that our some TV, which would be the distance between the Earth and Venus, would be 150 minus 108 times 10 to the six kilometers. And this is giving us 4.2 times 10 to the 10th meters. Ah, we can say the the distance between Earth and Jupiter would be equal to 778 minus 1 50 times 10 to the six kilometers. This is giving us 6.28 times 10 to the 11th meters. The distance between Earth and Saturn's would be equal to 1430 minus 150 times 10 to the six kilometers. And this is giving us 1.28 times 10 to the 12th meters. We can calculate the net force on the Earth by calculating the force of the gravitational force of Jupiter on the Earth. The gravitational force of Saturn on the earth minus the gravitational force of Venus on the earth. And that is because Venus would be technically if all of them are aligned with the sun, we can see the Venus would be on the other side of earth. So while Jupiter and Saturn would be closer to the outer solar system, we would be considering thie positive ex direction towards thie. Ah, out towards the end of the solar system, rather than towards the sun towards the sun would be considered the negative X direction. So we can say that this is going to be equal to the gravitational, constant times the mass of the earth squared and then times 318. This would be how many times more massive the planet is compared to Earth. So Jupiter has 318 times more mass than earth and then divided by the radius between Earth and Jupiter squared plus 95.1, divided by the radius between Earth and Saturn squared minus 0.815 divided by the distance between Earth and Venus squared my apologies. And so we can say that we can solve the net. Gravitational force on the earth will be 6.67 times 10 to the negative 11th times, 5.97 times 10 to the 24th kilograms Quantity squared and then this would be ah, 6.28 times 10 to the 11th meters squared on three eighteen 18 plus 95.1, given by 1.28 times 10 to the 12 meters quantity squared minus 0.815 divided by 4.2 times 10 to the 10th meters Quantity squared and we find that the net gravitational pull on the earth is going to be equal to 9.56 times 10 to the 17th Nunes. So this would be the answer to the first part. This would be the net gravitational pull of all the planets on earth. And then we want to find the ah net the gravitational pull of the sun on the Earth. Rather, we concise son on the earth and this would be equal to the gravitational, constant times the mass of the earth mass of the sun. But about the distance between the earth and the sun squared, this would be equal to 6.67 times 10 to the negative 11th times the mass of the Earth 5.97 times 10 to the 24th kilograms times 1.99 times 10 to the 30th kilograms on. Then this would be all divided by the radius of the rather the distance between the earth and the sun. And this is ah, 1.5 times 10 till the 11th meters quantity squared on. We find that the force of the sun on the earth would be equal Tio 3.552 times 10 to the 22nd Nunes s o Take note of this. The final part of the question asked us to find the ratio between these two. So the gravitational force of all the planets on earth, divided by the gravitational force of the sun on the earth, would be equal to 9.56 times 10 to the 17th divided by 3.552 times 10 to the 22nd and this is giving us 2.7 times 10 to the negative fifth. So essentially, this would be 27 millions. So then that gravitational force of all of all of the major planets on earth would only be 27 millions of the magnitude of the gravitational force of the sun on the earth. So extremely small. Yeah, the earth. The sun is, of course, what keeps us in our rotation. That is the end of the solution. Thank you for watching.

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