![Astronomy](https://www.bartleby.com/isbn_cover_images/9781938168284/9781938168284_largeCoverImage.gif)
What are the moons of the outer planets made of, and how is their composition different from that of our Moon?
![Check Mark](/static/check-mark.png)
The composition of the moons of the outer planets and the difference between their composition and our Moon’s composition.
Explanation of Solution
Introduction:
The first four planets of our solar system nearest to the Sun, that is, Mercury, Venus, Earth, and Mars lies inside the asteroid belt and are called the inner planets. The rest of the planets lie outside the asteroid belt and are called the outer planets. Moons are the natural satellites which revolve around the planets. The temperature of the outer planets and their moons is relatively very low due to the greater distance from the Sun.
The moons of the outer planets are composed of ice and rocks as their temperature is significantly low and due to this, the ice formation takes place easily. This ice is available in large quantity as the building material for the moons of the outer planets.
The composition of the moons of the outer planets differ from that of our Moon as the moons of outer planets are mainly composed of ice and rocks, whereas our Moon is mainly composed of rock only. Our Moon is situated closer to the Sun due to which, it does not have enough ice formation on it.
Conclusion:
Thus, the moons of the outer planets are composed of ice and rocks, whereas our Moon is mainly composed of rock only.
Want to see more full solutions like this?
Chapter 12 Solutions
Astronomy
Additional Science Textbook Solutions
Applications and Investigations in Earth Science (9th Edition)
Biology: Life on Earth (11th Edition)
Cosmic Perspective Fundamentals
Campbell Essential Biology with Physiology (5th Edition)
Campbell Essential Biology (7th Edition)
Physics for Scientists and Engineers: A Strategic Approach, Vol. 1 (Chs 1-21) (4th Edition)
- Problem Eight. A snowmobile is originally at the point with position vector 31.1 m at 95.5° counterclockwise from the x-axis, moving with velocity 4.89 m/s at 40.0°. It moves with constant acceleration 1.73 m/s² at 200°. After 5.00 s have elapsed, find the following. 9.) The velocity vector in m/s. (A)=-4.38+0.185ĵ (D) = 0.185 +4.38ĵ (B)=0.1851-4.38ĵ (E) = 4.38 +0.185ĵ (C) v=-0.1851-4.38ĵ (A)=-39.3-4.30ĵ 10.) The final position vector in meters. (B)=39.3-4.30ĵ (C) = -4.61 +39.3ĵ (D) = 39.31 +4.30ĵ (E) = 4.30 +39.3ĵarrow_forwardProblem Seven. A football receiver running straight downfield at 5.60 m/s is 11.5 m in front of the quarterback when a pass is thrown downfield at an angle of 35.0° above the horizon. 8.) If the receiver never changes speed and the ball is caught at the same height from which it was thrown, find the distance between the quarterback and the receiver when the catch is made. (A) 21.3 (B) 17.8 (C) 18.8 (D) 19.9 (E) 67.5arrow_forward3 Consider a ball sliding down a ramp as shown above. The ball is already in motion at the position 1. Which direction best approximates the direction of instantaneous velocity vector V when the object is at position 3?arrow_forward
- No chatgpt plsarrow_forwardA car in a roller coaster moves along a track that consists of a sequence of ups and downs. Let the x axis be parallel to the ground and the positive y axis point upward. In the time interval from t 0 tot = = 4s, the trajectory of the car along a certain section of the track is given by 7 = A(1 m/s)ti + A [(1 m/s³) t³ - 6(1 m/s²)t²]ĵ where A is a positive dimensionless constant. At t car ascending or descending? = 2.0 S is the roller coaster Ascending. Descending.arrow_forwardneed help on first part its not 220arrow_forward
- No chatgpt pls will upvotearrow_forwardNo chatgpt plsarrow_forwardChildren playing in a playground on the flat roof of a city school lose their ball to the parking lot below. One of the teachers kicks the ball back up to the children as shown in the figure below. The playground is 6.10 m above the parking lot, and the school building's vertical wall is h = 7.40 m high, forming a 1.30 m high railing around the playground. The ball is launched at an angle of 8 = 53.0° above the horizontal at a point d = 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (Due to the nature of this problem, do not use rounded intermediate values-including answers submitted in WebAssign-in your calculations.) (a) Find the speed (in m/s) at which the ball was launched. 18.1 m/s (b) Find the vertical distance (in m) by which the ball clears the wall. 0.73 ✓ m (c) Find the horizontal distance (in m) from the wall to the point on the roof where the ball lands. 2.68 m (d) What If? If the teacher always launches the ball…arrow_forward
- It is not possible to see very small objects, such as viruses, using an ordinary light microscope. An electron microscope can view such objects using an electron beam instead of a light beam. Electron microscopy has proved invaluable for investigations of viruses, cell membranes and subcellular structures, bacterial surfaces, visual receptors, chloroplasts, and the contractile properties of muscles. The "lenses" of an electron microscope consist of electric and magnetic fields that control the electron beam. As an example of the manipulation of an electron beam, consider an electron traveling away from the origin along the x axis in the xy plane with initial velocity ₁ = vi. As it passes through the region x = 0 to x=d, the electron experiences acceleration a = ai +a, where a and a, are constants. For the case v, = 1.67 x 107 m/s, ax = 8.51 x 1014 m/s², and a = 1.50 x 10¹5 m/s², determine the following at x = d = 0.0100 m. (a) the position of the electron y, = 2.60e1014 m (b) the…arrow_forwardNo chatgpt plsarrow_forwardneed help with the first partarrow_forward
- AstronomyPhysicsISBN:9781938168284Author:Andrew Fraknoi; David Morrison; Sidney C. WolffPublisher:OpenStaxFoundations of Astronomy (MindTap Course List)PhysicsISBN:9781337399920Author:Michael A. Seeds, Dana BackmanPublisher:Cengage Learning
- An Introduction to Physical SciencePhysicsISBN:9781305079137Author:James Shipman, Jerry D. Wilson, Charles A. Higgins, Omar TorresPublisher:Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781938168284/9781938168284_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781337399920/9781337399920_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781337672252/9781337672252_smallCoverImage.jpg)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305079137/9781305079137_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781305804562/9781305804562_smallCoverImage.jpg)