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Kepler’s second law of planetary motion describes the area “swept out” by the line connecting a planet and the sun during equal time intervals. The conceptual message of the law is that ______.
Planets move faster when they are near the sun, and move slower when they are farther away
It accurately reflects Kepler's second law by emphasizing the variation in orbital speed.
It explains that the equal areas are swept out despite the changing speed.
It directly states the conceptual message of the law.
Kepler’s first law of planetary motion tells us that planets move along elliptical orbits. If we picture the elliptical orbit of a planet orbiting the sun, where does the sun fit into the picture?
The sun is at one focus of the ellipse, with nothing at the other focus.
This choice accurately describes the structure of an ellipse in planetary orbits.
The ellipse has two foci; one is occupied by the sun, and the other remains empty.
This aligns with Kepler's first law, where the sun occupies one focus.
A distant solar system has a star ("Alpha") that is similar to our sun, but it has only half the mass of our sun. In addition, the other solar system has a planet ("Planet X") just like the earth, with the same mass and the same distance from its star as the earth is from our sun. How long does it take Planet X to orbit Alpha once?
516 days EXPLANATION:
Step 1:
Write down Kepler’s Third Law for orbital motion. It is given by: T² = (4π^2 a^3) / GM where T is the orbital period, a is the orbital radius, G is the gravitational constant, and M is the mass of the star.
Step 2:
For Earth orbiting our sun, we have: T_Earth² = (4π^2 a^3) / GM and we know that T_Earth = 1 year
Step 3:
For Planet X orbiting the star Alpha with mass M_α = 0.5 M⊙, the orbital period becomes:
T_X² = (4π^2 a^3) / G(0.5 M⊙) = (4π^2 a^3) / (0.5 GM⊙) = 2⋅(4π^2 a^3) / (GM⊙) = 2T_Earth²
Step 4:
Taking the square root of both sides gives:
T_X = √2⋅T_Earth = √2⋅1 year
Step 5:
Evaluating √2 approximately as 1.4142, we find:
T_X ≈ 1.4142 years
(r_A + d + r_B)² EXPLANATION:
Step 1: Identify the radii of the spheres. Sphere A has a radius r_A and Sphere B has a radius r_B.
Step 2: Determine the distance between the centers of the two spheres. This distance is the sum of the radii of the two spheres plus the distance between their surfaces. So, the total distance is r_A + r_B + d.
Step 3: Recall the formula for Newton's law of universal gravitation:
F = G ((m_1m_2) / r² )
where F is the gravitational force, G is the gravitational constant, my and me are the masses of the two objects, and r is the distance between the centers of the two masses.
Step 4: Substitute the total distance from Step 2 into the denominator of the formula. So, the denominator becomes (r_a+ r_b + d)².
Step 5: Use this expression in the formula to calculate the gravitational force between the two spheres. Remember, the key is to use the correct distance between the centers of the spheres!
A ball is dropped from a height above the earth's surface equal to 𝑅, the radius of the earth (so the ball starts off a distance 2𝑅 from the center of the earth). The ball falls under the influence of gravity until it hits the surface. Assume no energy is lost to air resistance.
What is the speed of the ball by the time it gets to the earth's surface?
For reference:
𝐺 = 6.67 x 10-11 Nm2/kg2 (gravitational constant)
𝑀 = 5.97 x 1024 kg (mass of earth)
𝑅 = 6.37 x 106 m (radius of earth)
7.91×10³ m/s INSERT LINER EXPLANATION:
Step 1:
Express the gravitational potential energy as U(r) = −GMm / r. The ball starts with potential energy at r_i = 2R and ends with potential energy at r_f = R. Its kinetic energy starts at zero (dropped from rest) and increases as it falls.
Step 2:
Write the conservation of energy equation between the initial and final states: −GMm / 2R = ½ mv² − GMm / R
Step 3:
Solve for the kinetic energy at impact. Rearranging the energy conservation equation:
½ mv² = −GMm / 2R + GMm/ R = GMm / 2R
Cancel the mass (m) as ∑∈g (m ≠ 0):
[½ v² = GM / 2R] Multiply both sides by 2: \\[ v^2 = \frac{GM}{R} \\] Thus, the speed \\(v\\) is: \\[ v = \sqrt{\frac{GM}{R}} \\] Step 4 Substitute the given values: \\[ G = 6.67 \times 10^{-11} ~\\mathrm{Nm^2/kg^2}, \\quad M = 5.97 \times 10^{24} ~\\mathrm{kg}, \\quad R = 6.37 \times 10^{6} ~\\mathrm{m} \\] Compute the product \\(GM\\): \\[ GM = (6.67 \times 10^{-11}) (5.97 \times 10^{24}) \\approx 3.9860 \times 10^{14} ~\\mathrm{m^3/s^2} \\] Now, compute \\(v\\): \\[ v = \sqrt{\\frac{3.9860 \times 10^{14}}{6.37 \times 10^{6}}} \\approx \\sqrt{6.2600 \times 10^{7}} ~\\mathrm{m/s} \\] Taking the square root: \\[ v \\approx 7.9125 \\times 10^{3} ~\\mathrm{m/s} \\] Step 5 Round the result appropriately, maintaining four decimal place precision where applicable. <finalAnswer> 7.9125 x 10^3 m/s </finalAnswer> FINISH LATER
Newton’s law of universal gravitation states that the magnitude of the force of gravity between you and the earth gets smaller when you get farther away from the center of the earth. Why is it that we use the same magnitude of the acceleration due to gravity (g = 9.8 m/s2) for all physics problems without paying attention to whether the problem takes place at sea level or at the top of a tall mountain?
The height of a tall mountain is a tiny amount compared to the radius of the earth. So even going to the top of a tall mountain barely changes your distance to the center of the earth. INSERT LINER EXPLANATION:
This choice is correct because the variation in altitude when going from sea level to the top of a tall mountain is extremely small relative to the Earth's radius.
The gravitational acceleration, g = GM / r² changes very little because the difference in r is negligible.
A mountain’s height (a few kilometers) is much smaller compared to the Earth's radius (approximately 6.37 × 10^6, m), so the change in g is minimal.
The magnitude of the force of gravity between objects A and B is 𝐹. If the mass of object A was twice as large as it is, but if everything else was kept the same, what would be the new magnitude of the force of gravity between objects A and B?
2F INSERT LINER EXPLANATION: Step 1 Identify the original formula for the gravitational force: F = G m A m B r 2 F=G r 2 m A m B where G G is the gravitational constant, m A m A and m B m B are the masses of objects A and B, respectively, and r r is the distance between their centers of mass. Step 2 Double the mass of object A, so it becomes 2 m A 2m A . Substitute this back into the equation for the force of gravity: F n e w = G ( 2 m A ) m B r 2 = 2 ( G m A m B r 2 ) F new =G r 2 (2m A )m B =2(G r 2 m A m B ) Step 3 Recognize that the expression G m A m B r 2 G r 2 m A m B is the original force F F. Therefore, the new force becomes: F n e w = 2 F F new =2F
The magnitude of the net force on each object is the same. INSERT PEARSON EXPLANATION
Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that has the same radius as earth, but is about 10% less massive. Roughly, what acceleration magnitude due to gravity would you expect if you were standing on the surface of this new planet?
Roughly 8.8 m/s². INSERT LINER EXPLANATION
Exoplanets (planets outside our solar system) are an active area of modern research. Suppose you read an article stating that there is a newly discovered planetary system with three planets. The article states that the outermost planet (Planet C) goes all the way around its star in less time than the innermost planet (Planet A). According to Kepler’s laws of planetary motion, is this possible?
This is not possible, since it would violate Kepler’s third law of planetary motion. INSERT LINER EXPLANATION
According to Kepler’s third law of planetary motion, if the earth’s distance from the sun were twice as large, how long would it take for the earth to go around the sun once?
About 2.8 times as long INSERT LINER EXPLANATION
Two masses are at rest a distance 𝑑 apart from one another. An external agent moves one mass so that it is a distance 2𝑑 from the other mass. Which of the following is true?
The external agent did positive work on the two-mass system. INSERT LINER EXPLANATION
Currently, the moon takes 27.3 days to go around the earth. Suppose a super villain shifted the moon into a new, lower, circular orbit. How long would it take for the moon to go around the earth in its new circular orbit?
Less than 27.3 days INSERT LINER EXPLANATION
A ball of mass 200 kg is dropped from a height above the earth's surface equal to 𝑅, the radius of the earth (so the ball starts off a distance 2𝑅 from the center of the earth). The ball falls under the influence of gravity until it hits the surface. We will not ignore air resistance in this problem.
If the ball only has a speed of 2.41 x 103 m/s when it hits the earth's surface, how much energy was dissipated by air resistance?
For reference:
𝐺 = 6.67 x 10-11 Nm2/kg2 (gravitational constant)
𝑀 = 5.97 x 1024 kg (mass of earth)
𝑅 = 6.37 x 106 m (radius of earth)
5.67×109 J
Kepler’s third law of planetary motion states that which two quantities are related?
A planet’s orbital period and the semimajor-axis length are related. INSERT LINER EXPLANATION OR ARC SEARCH
If the (earth+sun) potential energy is 𝑈, and the kinetic energy of the earth as it moves around the sun is 𝐾, then which of the following best describes the relationship between 𝐾 and 𝑈?
𝐾 = −𝑈/2 INSERT LINER EXPLANATION
A communications company requests that their satellite be placed 100 km above the earth's surface and then launched so that it moves in a geosynchronous orbit (meaning an orbit that takes exactly 24 hours to complete). Is this possible?
No, orbiting every 24 hours only happens at one specific orbital distance. INSTER LINER EXPLANATION
The magnitude of the force of gravity on object B by object A is 𝐹. If the distance between the objects was tripled, but everything else was kept the same, what would be the new magnitude of the force of gravity on object B by object A?
𝐹/9 INSTER LINER EXPLANATION
Which of the following best describes the best way for a satellite to transfer from its current circular orbit around the earth to another circular orbit with a larger radius?
First have the thrust give a boost in the direction the satellite is moving, then later have another thrust giving another boost in the direction the satellite is moving. INSERT LINER EXPLANATION
Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that has the same mass as earth, but has a radius that is about 10% less. Roughly, what acceleration magnitude due to gravity would you expect if you were standing on the surface of this new planet?
Roughly 12.1 m/s2
INSERT LINER EXPLANATION
The magnitude of the acceleration due to gravity on the surface of the moon is approximately 0.17𝑔 (where 𝑔 is the value on the surface of the earth). If the moon's radius is 27% the earth's radius, then what is the approximate mass of the moon in terms of the earth's mass, 𝑀
0.012 𝑀
If a planet is orbiting a star, what is certainly true regarding the shape of the orbit?
The orbit is shaped like an ellipse. INSERT LINER EXPLANATION
An object is moved from the surface of the earth to a height above the surface equal to the radius of the earth. Which of the following is true about the potential energy of the (earth + object) system? There may be more than one correct choice.
The potential energy increases and decreases in absolute value. EXPLANATION: The potential energy is 𝑈 = −𝐺𝑀𝑚 / 𝑟. As 𝑟 increases, then the absolute value of 𝑈 decreases, but the value of 𝑈 increases (it remains negative, increases, and becomes smaller in absolute value).
A rocket is fired vertically upward with half the escape speed. Assuming no energy is lost to air resistance, find how high above the earth's surface the rocket will reach in terms of the earth's radius, 𝑅.
𝑅 / 3 INSERT LINER EXPLANATION
