/*! 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} Q33 PE (a) Calculate Earth’s mass giv... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Calculate Earth’s mass given the acceleration due to gravity at the North Pole is 9.830 m/s2and the radius of the Earth is 6371kmfrom pole to pole. (b) Compare this with the accepted value ofM=5.979×1024kg.

Short Answer

Expert verified

(a) The calculated mass of the earth isM=5.979×1024kg

(b) The calculated value is matching with accepted value.

Step by step solution

01

Definition of Gravity

Gravity is a universal phenomenon and is introduced by Newton and Derived the expression for gravitational force.

02

Calculating the mass of Earth

Earth’s radius= R = 6371 x 103m

Gravitational acceleration near the North Pole= a = 9.83 m/s2

Earth’s mass= M = ?

The gravitational force is calculated as follows:

F = GMmR2

Here, mis object’s mass, Mis Earth’s mass, R Earth’s radius is the gravitation constant and F is the gravitational force. The force acting on the object is calculated as follows:F=ma, here, is the object’s acceleration.

ma = GMmR2

M = aR2/G

M=9.83×6371×1032/6.637×10-11

M=5.979×1024kg

This is the calculated mass of the earth.

03

Determining difference between the calculated and accepted value

The calculated value of mass is M=5.979×1024kgwhich is exactly matching with the accepted value of the mass of the earth. The difference in percent between the computed and approved values is zero percent.

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 91Ó°ÊÓ!

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

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads). (a) Calculate the ideal speed to take a \(100{\rm{ m}}\) radius curve banked at \(15.0^\circ \). (b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at \(20.0{\rm{ km}}/{\rm{h}}\)?

Unreasonable Results (a) Calculate the minimum coefficient of friction needed for a car to negotiate an unbanked \({\bf{50}}.{\bf{0}}{\rm{ }}{\bf{m}}\) radius curve at\({\bf{30}}.{\bf{0}}{\rm{ }}{\bf{m}}/{\bf{s}}\). (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent?

Do you feel yourself thrown to either side when you negotiate a curve that is ideally banked for your car’s speed? What is the direction of the force exerted on you by the car seat?

(a) What is the acceleration due to gravity on the surface of the Moon?

(b) On the surface of Mars? The mass of Mars is \({\bf{6}}.{\bf{418}} \times {\bf{1}}{{\bf{0}}^{{\bf{23}}}}{\bf{kg}}\) and its radius is\({\bf{3}}.{\bf{38}} \times {\bf{1}}{{\bf{0}}^{\bf{6}}}{\bf{m}}\).

When a toilet is flushed or a sink is drained, the water (and other material) begins to rotate about the drain on the way down. Assuming no initial rotation and a flow initially directly straight toward the drain, explain what causes the rotation and which direction it has in the northern hemisphere. (Note that this is a small effect and in most toilets the rotation is caused by directional water jets.) Would the direction of rotation reverse if water were forced up the drain?

See all solutions

Recommended explanations on Physics Textbooks

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.