Calculating Energy for Orbital Deployment

In summary, the scientist needs to determine the work required to put a 100kg experimental package into orbit around the Earth at a height of 1,000km. This involves finding the potential energy and the work done by gravity, taking into account that gravity decreases with height. The amount of extra work needed to put the package into a circular orbit at this height must also be calculated.
  • #1
ashvuck101
27
0

Homework Statement



A scientist wants to put an 100kg experimental package in orbit around the Earth. The cost of deployment depends on the amount of extra energy it takes to get it into the required position i.e. how much more energy is used than just sending the rocket up there.

a) Determine the amount of work that must be done to get the package to 1 000km above the Earth’s surface.

b) Determine the amount of extra work needed to put the package into a circular orbit at this height.


Homework Equations



Ep=-Gm1m2/r

w=fd

f=Gm1m2/r2

The Attempt at a Solution



using

f=Gm1m2/r2

= 9.83*102

w=fd
= 9.83*102*100000
= 9.83*107


b)

im not sure if a is right and i don't know how to get the answer for b

somebody at least tell me i am going in the right direction here
 
Physics news on Phys.org
  • #2
Hi ashvuck101,

ashvuck101 said:

Homework Statement



A scientist wants to put an 100kg experimental package in orbit around the Earth. The cost of deployment depends on the amount of extra energy it takes to get it into the required position i.e. how much more energy is used than just sending the rocket up there.

a) Determine the amount of work that must be done to get the package to 1 000km above the Earth’s surface.

b) Determine the amount of extra work needed to put the package into a circular orbit at this height.


Homework Equations



Ep=-Gm1m2/r

w=fd

f=Gm1m2/r2

The Attempt at a Solution



using

f=Gm1m2/r2

= 9.83*102

Okay, this is the force of gravity on the object at the Earth's surface.

w=fd
= 9.83*102*100000

You seem to be missing a zero here for the distance.

But more importantly, this approach is what you would use if the gravitational force were constant over the entire 1000 km. (Because to find the minimum energy, you would set the pushing force to be equal to the gravitational force.) However, gravity decreases with height. How do you find the work of a force that is not constant?

As an alternative, what is the definition of potential energy? How can that help you find the work done by gravity?


= 9.83*107


b)

im not sure if a is right and i don't know how to get the answer for b

somebody at least tell me i am going in the right direction here
 

Related to Calculating Energy for Orbital Deployment

1. What are orbitals and how do they work?

Orbitals are the three-dimensional regions around an atom's nucleus where electrons are most likely to be found. They work by following the laws of quantum mechanics, which dictate the behavior of particles at the atomic level. Electrons are able to move within the orbital and stay in a stable orbit due to the balance between their kinetic energy and the attractive force from the nucleus.

2. How are orbitals different from orbits?

Orbits refer to the path of an object around a larger object, such as a planet around a star. Orbitals, on the other hand, are specific regions within an atom where electrons can exist. Unlike orbits, which have a defined path, orbitals are more like a cloud of probability, showing where an electron is most likely to be found at any given time.

3. What is the significance of escape velocity?

Escape velocity is the minimum velocity needed for an object to break free from the gravitational pull of a larger object. In terms of orbitals, escape velocity is important because it determines whether an electron will be able to escape from the atom's nucleus and become a free electron. This plays a crucial role in chemical reactions and the behavior of matter.

4. How is escape velocity calculated?

The formula for escape velocity is: v = √(2GM/r), where v is the escape velocity, G is the gravitational constant, M is the mass of the larger object, and r is the distance between the two objects. This formula shows that escape velocity is dependent on the mass and distance of the objects involved.

5. Can escape velocity be exceeded?

Yes, it is possible for an object to exceed escape velocity. When an object exceeds escape velocity, it will leave its current orbit and travel on a new path, either towards or away from the larger object. This is known as an escape trajectory and is often used in space missions to explore other planets and celestial bodies.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
2
Replies
39
Views
3K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
941
  • Introductory Physics Homework Help
Replies
2
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top