The orbital period, average distance from the Sun and orbital velocity of asteroid.
Answer to Problem 7P
The orbital period of the asteroid is
Explanation of Solution
Write the expression for the orbital period of the asteroid.
Here,
Write the expression for the Kepler’s third law.
Here,
Write the expression for the orbital velocity of the asteroid.
Here,
Refer to the Celestial Profile of Jupiter to obtain the orbital period of Jupiter as
Conclusion:
Substitute
Substitute
Substitute
Therefore, the orbital period of the asteroid is
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Chapter 24 Solutions
FOUNDATIONS OF ASTRON.-MINDTAP (2 TERM)
- Dione, a moon of Saturn, has an orbital radius of 377,400 km, and an orbital period of about 2.737 Earth days. Find the orbital period of Rhea, another moon of Saturn, which has an orbital radius of 527,040 km. Find the period in Earth days. Round to the nearest hundredth. Don't worry about putting the unit, just put the answer.arrow_forwardI. Directions: Complete the given table by finding the ratio of the planet's time of revolution to its radius. Average Radius of Orbit Times of Planet R3 T2 T?/R3 Revolution Mercury 5.7869 x 1010 7.605 x 106 Venus 1.081 x 1011 1.941 x 107 Earth 1.496 x 1011 3.156 x 107 1. What pattern do you observe in the last column of data? Which law of Kepler's does this seem to support? II. Solve the given problems. Write your solution on the space provided before each number. 1. You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth's surface. Find the following: a) Speed b) Period c) Radial Acceleration Given: Unknown: Formula: Solution: Answer: Given: Unknown: Formula: Solution: Answer: Given: Unknown: Formula: Solution: Answer:arrow_forwardHow long (in Earth-Years) does it take Saturn to orbit the Sun? Use these values of (average) distance to the Sun. Venus: .72 A.U. Saturn: 9.5 A.U. Give your answer in (Earth) years to the correct number of significant figures.arrow_forward
- The value we have just calculated is the combined masses of Jupiter and Callisto! Their mass is in units of the Sun's Mass (MS) - we can convert this to units which are more familiar to us like kilograms by multiplying this answer by the scale factor (1.99x1030 kg/1 MS): (MJupiter + MCallisto) = ( MS) (1.99x1030 kg/1 Solar Mass) = _______x_10___ kg (I have already written the x 10 so you are reminded to write the exponenet of 10 in the scientific notation expression of your answer). Since you know from looking at pictures of Jupiter with its Galilean Satellites (look in your book at a picture if you have not already), that Callisto is much smaller than Jupiter - in fact it is less than 0.001 of Jupiter's size or mass, so the number we have just calculated for (MJupiter + MCallisto) is almost the same as MJupiter . How much more massive is Jupiter than the Earth? (The mass of Earth is about 5.98 x 1024 kg)arrow_forwardUse the small-angle formula to calculate the angular diameter (in arc minutes) of Mars (d = 6.79 ✕ 103 km) as seen from Earth if Mars were at the location of the Sun (D = 1.5 ✕ 108 km).arrow_forwardList some reasons that the study of the planets has progressed more in the past few decades than any other branch of astronomy.arrow_forward
- How Do We Know? How can a scientific model be useful if it is not a true description of nature?arrow_forwardGalileos telescope showed him that Venus has a large angular diameter (61 arc seconds) when it is a crescent and a small angular diameter (10 arc seconds) when it is nearly full. Use the small-angle formula to find the ratio of its maximum to minimum distance from Earth. Is this ratio compatible with the Ptolemaic universe shown in Figure 3b of the Chapter 4 Concept Art: An Ancient Model of the Universe?arrow_forwardAssume that the planet's orbit is circular of radius R = 130 × 106 km and planet's period is T = 30 × 10° s. What is the magnitude of the vector J = r x r' (in units of square kilometers per second)? (Use decimal notation. Give your answer to three decimal places.) ||J|| = x10° km²/s Find the rate at which the planet's radial vector sweeps out area in units of square kilometers per second. (Use decimal notation. Give your answer to three decimal places.) dA x10° km²/s dtarrow_forward
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