/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q56P Your starship, the Aimless Wande... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Q56P (page 428)

Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: A 2.50-kg stone thrown upward from the ground at 12.0 m/s returns to the ground in 4.80 s; the circumference of Mongo at the equator is 2.00 x 105 km; and there is no appreciable atmosphere on Mongo. The star ship commander, Captain Confusion, asks for the following information: (a) what is the mass of Mongo? (b) If the Aimless Wanderer goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?

Short Answer

Expert verified

a) Mass of the mango M=8.778x1025kg

b) Time in hours T = 11.08h

Step by step solution

01

Identification of the given data

Circumference of mango C=2脳105km=2.0脳108m

Time t = 4.80 s

Mass of stone M = 2.50 kg

Velocity role="math" localid="1668068026698" v1=12m/s

Altitude h=3脳107m

02

Calculation for mass of mango

As we know that

v2=v1t鈭�12gt20=13脳4.50鈭�12脳g脳4.502g=5.78m/s2

Newton鈥檚 law of gravity is given by

onclusion statement

M=gC24蟺2GM=5.78脳2.0脳10824蟺2脳6.672脳10鈭�11M=8.778脳1025kg
03

Calculation for time

As we know Kepler鈥檚 third law,

T=2蟺C2蟺+h3(G脳M)12T=2蟺2.0脳1082蟺+3脳107326.672脳10鈭�11脳8.778脳102512T=39917.5sT=11.08h

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 950-kg cylindrical can buoy floats vertically in seawater. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0-kg man stands on top of it.

You are on the roof of the physics building, 46.0鈥�mabove the ground (Fig. P2.70). Your physics professor, who is1.80鈥�mtall, is walking alongside the building at a constant speed of1.20鈥�m/s. If you wish to drop an egg on your professor鈥檚 head, where should the professor be when you release the egg? Assume that the egg is in free fall.

(Fig. P2.70)

Given two vectorsA鈫�=鈭�2.00i^+3.00j^+4.00k^andB鈫�=3.00i^+1.00j^鈭�3.00k^, (a) find the magnitude of each vector; (b) use unit vectors to write an expression for the vector differenceA鈫�鈭�B鈫�; and (c) find the magnitude of the vector differenceA鈫�鈭�B鈫�. Is this the same as the magnitude ofB鈫�鈭�A鈫�? Explain.

A circular racetrack has a radius of 500 m. What is the displacement of a bicyclist when she travels around the track from the north side to the south side? When she makes one complete circle around the track? Explain.

A car is stopped at a traffic light. It then travels along a straight road such that its distance from the light is given byxt=bt2-ct3, whereb=2.40m/s2andc=0.120m/s3. (a) Calculate the average velocity of the car for the time interval t = 0 to t = 10.0 s. (b) Calculate the instantaneous velocity of the car at t = 0, t = 5.0 s, and t = 10.0 s. (c) How long after starting from rest is the car again at rest?
See all solutions

Recommended explanations on Physics Textbooks

Wave Optics

Read Explanation

Work Energy and Power

Read Explanation

Astrophysics

Read Explanation

Solid State Physics

Read Explanation

Oscillations

Read Explanation

Nuclear Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.