<< Chapter < Page Chapter >> Page >

Lifting a payload

How much energy is required to lift the 9000-kg Soyuz vehicle from Earth’s surface to the height of the ISS, 400 km above the surface?

Strategy

Use [link] to find the change in potential energy of the payload. That amount of work or energy must be supplied to lift the payload.

Solution

Paying attention to the fact that we start at Earth’s surface and end at 400 km above the surface, the change in U is

Δ U = U orbit U Earth = G M E m R E + 400 km ( G M E m R E ) .

We insert the values

m = 9000 kg, M E = 5.96 × 10 24 kg, R E = 6.37 × 10 6 m

and convert 400 km into 4.00 × 10 5 m . We find Δ U = 3.32 × 10 10 J . It is positive, indicating an increase in potential energy, as we would expect.

Significance

For perspective, consider that the average US household energy use in 2013 was 909 kWh per month. That is energy of

909 kWh × 1000 W/kW × 3600 s/h = 3.27 × 10 9 J per month.

So our result is an energy expenditure equivalent to 10 months. But this is just the energy needed to raise the payload 400 km. If we want the Soyuz to be in orbit so it can rendezvous with the ISS and not just fall back to Earth, it needs a lot of kinetic energy. As we see in the next section, that kinetic energy is about five times that of Δ U . In addition, far more energy is expended lifting the propulsion system itself. Space travel is not cheap.

Got questions? Get instant answers now!

Check Your Understanding Why not use the simpler expression Δ U = m g ( y 2 y 1 ) ? How significant would the error be? (Recall the previous result, in [link] , that the value g at 400 km above the Earth is 8.67 m/s 2 .)

The value of g drops by about 10% over this change in height. So Δ U = m g ( y 2 y 1 ) will give too large a value. If we use g = 9.80 m/s , then we get

Δ U = m g ( y 2 y 1 ) = 3.53 × 10 10 J

which is about 6% greater than that found with the correct method.

Got questions? Get instant answers now!

Conservation of energy

In Potential Energy and Conservation of Energy , we described how to apply conservation of energy for systems with conservative forces. We were able to solve many problems, particularly those involving gravity, more simply using conservation of energy. Those principles and problem-solving strategies apply equally well here. The only change is to place the new expression for potential energy into the conservation of energy equation, E = K 1 + U 1 = K 2 + U 2 .

1 2 m v 1 2 G M m r 1 = 1 2 m v 2 2 G M m r 2

Note that we use M , rather than M E , as a reminder that we are not restricted to problems involving Earth. However, we still assume that m < < M . (For problems in which this is not true, we need to include the kinetic energy of both masses and use conservation of momentum to relate the velocities to each other. But the principle remains the same.)

Escape velocity

Escape velocity is often defined to be the minimum initial velocity of an object that is required to escape the surface of a planet (or any large body like a moon) and never return. As usual, we assume no energy lost to an atmosphere, should there be any.

Consider the case where an object is launched from the surface of a planet with an initial velocity directed away from the planet. With the minimum velocity needed to escape, the object would just come to rest infinitely far away, that is, the object gives up the last of its kinetic energy just as it reaches infinity, where the force of gravity becomes zero. Since U 0 as r , this means the total energy is zero. Thus, we find the escape velocity    from the surface of an astronomical body of mass M and radius R by setting the total energy equal to zero. At the surface of the body, the object is located at r 1 = R and it has escape velocity v 1 = v esc . It reaches r 2 = with velocity v 2 = 0 . Substituting into [link] , we have

Questions & Answers

Quantity which are used in physics simply
Sangram Reply
That's philosophical question.
Jan
What is the Physical quantity
Raja Reply
What is Centripetal force
Taiwo Reply
a force of attraction that tends to keep a body moving in a circular path
Mustapha
force pulling the particle towards the center when moving in circular path
juny
yes
Mustapha
good job
Pranshu
y oxygen is shift to first place
Radhika Reply
What is coriolis firce
Shakeel Reply
Angular accelaration force,as the result of the rotatation of earth
Arzoodan
Due to the rotation of the earth the winds at the equator get deflected in opposite direction and therefore cause some currents in the Northern and Southern hemisphere which are opposite in there spin.
Varsha
Thats correct.
Arzoodan
if cos = x/y then sec = y/x if sin = y/x then cosec = x/y
Jan Reply
if cosβ=x/y then what is cosecβ
Abubakar Reply
. if cosβ=x/y what is cosecβ
Abubakar
if cosΦ=x/y what is cosecΦ
Abubakar
What are the unknown symbols?
Jan
β
Jan
Φ
Jan
It is defined that cosec = 1/sin and sec = 1/cos
Jan
Do you understand this?
Jan
of cours
Arzoodan
∇(f/g) = (g∇f − f∇g)/g^2 , at points x where g(x) 6= 0 please help me to solve the problem ....
11 Reply
sir please add answer sheet of spigel vector analysis ..
11 Reply
In a pulley system 2 boxes r hanging in both sides of pulley. An other box was joined to the left box through a rope & get accelerated downward. If all the boxes have same mass ,then what will be the acceleration of that system ?
bibek Reply
it may be mg
Amalesh
∇(f/g) = (g∇f − f∇g)/g2 , at points x where g(x) 6= 0 plus solve it
11
What is momentum
Rika Reply
is the product of mass to it's velocity (mv)
Ahmed
momentum
oladipupo
quatity of motion present in a body or product of mass and velocity
Shakeel
show that the kE of a uniform ring of mass m rolling along a smooth horizontal surface so that its centre of mass has a velocity v is mv×v
folder
what is hydration energy
osobase Reply
the energy................................................
Rika
what is momentum
Ogwu Reply
dont know want to know the answer
Rika
it's the product of mass multiplied by velocity
harsha
the quantity of motion possessed by a body is called its momentum.by virtue of which a body can exert a force in the agency which tend to stop it .it is a common experience that stronger force is required to stop more massive body.also faster the body moves harder it is to stop it .this is why
Manoj
momentum is product of mass and velocity and it is denoted by "P".
Manoj
show that the cross product of vector axb=-bxa
Teklu Reply
Nature of the physical low
Rashika Reply
1/2 mv2 =mgh prove that
Rashika
study work energy theorem properly
Rohan
v2=u2+2gh...u=0...so 1/2 mv2=1/2m(2gh)=mgh
Talha
Practice Key Terms 2

Get the best University physics vol... course in your pocket!





Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'University physics volume 1' conversation and receive update notifications?

Ask