Question

In each hand you hold a 0.17-kg apple. What is the gravitational force exerted by each apple on the other when their separation is each of the following?

(a) 0.40 m
_______ N

(b) 0.80 m
_______ N

A communications satellite with a mass of 500 kg is in a circular orbit about the Earth. The radius of the orbit is 46,000 km as measured from the center of the Earth.

(a) Calculate the weight of the satellite on the surface of the Earth.
______ kN

(b) Calculate the gravitational force exerted on the satellite by the Earth when it is in orbit.
______ kN

An 8.84-kg block slides with an initial speed of 1.54 m/s down a ramp inclined at an angle of 26.4

Answer 1

1. So Newton's Law of Gravitation yields F = Gm1m2/r^2
   
   So F = 6.67x10^-11N-m^2/kg^2*0.17 kg* 0.17 kg/(0.4^2) m^2 = 1.204 x10^-11N
   
   b) F = 6.67x10^-11N-m^2/kg^2*0.17 kg* 0.17 kg/(0.8^2) m^2 = 0.3012 x10^-11N

Part 2

1.) The mass doesn't change on the surface on the Earth. So on Earth the mass of the satellite is also 500 kg. The acceleration of gravity on the surface of the Earth is 9.81 m/s^2. So the weight is:
W = mg
W = (500 kg) * (9.81 m/s^2)
W = 4905 N
W = 4.905 kN

2.) The gravitational force is:
F = (GMm) / r^2
WHERE:
F = Force of gravity
G = Gravitational constant (6.673 x 10^-11 m^3 / (kg*s^2)
M = Mass of the planet (5.9742 x 10^24 kg for Earth)
m = mass of the satellite (500 kg)
r = distance from the center of the planet to the point in question (46,000,000 m)

F = (6.673E-11)*(5.9742E24)*(500 kg) / (46,000,000^2)
F = 94.093 N
F = 0.094 kN

Part 3

The change in kinetic energy is -1/2 m v^2.

The work done by gravity on the block is

Wgrav = - mg h = - m g d sin(angle)


The work done by friction on the block is

Wfric = -Ffric d = - mu mg cos(angle) d

The work-energy theorem equates change in KE to net work done:

-1/2 m v^2 = -mgd sin(angle) - mu mg d cos(angle)

Therefore

d = v^2/(2g (sin(angle)+mu cos(angle) )

d = (1.54m/s)^2 /(2*9.81m/s^2*(sin(26.4) +0.63 cos(26.4))

d= 0.1198 m = 11.98 cm