Everything Science Grade 11

CHAPTER 12. FORCE,MOMENTUM AND IMPULSE 12.4

Step 3 : Situation on Earth

wE = mgE= G

mE· m

rE^2

= (70kg)(9,8m· s−^2 )

= 686N

Step 4 : Situation on Zirgon interms of situation on Earth

Write the equation for the gravitational force onZirgon and then substitute the

values for mZand rZ, in terms of the values for the Earth.

wZ= mgZ = G

mZ· m

r^2 Z

= G

2 mE· m

r^2 E

= 2(G

mE· m

r^2 E

)

= 2wE

= 2(686N)

= 1372N

Step 5 : Quote the final answer

The man weighs 1 372N on Zirgon.

Example 25: Comparative Problem 2

QUESTION

A man has a mass of 70kg. On the planet Beeble how much will he weigh if Beeble has mass

half of that of the Earthand a radius one quarter that of the Earth. Gravitational acceleration

on Earth is 9,8 m·s−^2.

SOLUTION

Step 1 : Determine what information has been given

The following has beenprovided:

• the mass of the man onEarth, m


• the mass of the planetBeeble (mB) in terms of the mass ofthe Earth (mE), mB=^12 mE


• the radius of the planet Beeble (rB) in terms of the radius of the Earth (rE), rB=^14 rE

Step 2 : Determine how to approach the problem

We are required to determine the man’s weighton Beeble (wB). We can do this

by using:

w = mg = G

m 1 · m 2

r^2

(12.7)

to calculate the weight of the man on Earth andthen use this value to determine

the weight of the man on Beeble.