Circular Orbit / Practice Quiz #4, Question 27

In summary: Earth's surface= F=GMm/r^2 acceleration due to gravity at 1.00 x 10^6 above the Earth's surface= F=GMm/r^2 ratio of the equations= F/F=0
  • #1
gcombina
157
3

Homework Statement

What is the acceleration due to gravity at an altitude of 1.00 x 10^6 above the Earth's surface, given that the radius of the Earth is 6.38 x 10^6 m?

Homework Equations


Using g = GMm/r^2
Fc = mv^2/r

The Attempt at a Solution


Fc = mv^2/r

How can I solve this equation if I am missing m and v?
 
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  • #2
The only force acting upon the object in this example is the force of Earth's gravity. You only need to use the equation F=GMm/r^2...remember F=ma, which means that acceleration=G*M/r^2 and mass =m. F=m*a=m*(GM/r^2)
 
  • #3
gcombina said:

Homework Statement

What is the acceleration due to gravity at an altitude of 1.00 x 10^6 above the Earth's surface, given that the radius of the Earth is 6.38 x 10^6 m?

Homework Equations


Using g = GMm/r^2
Fc = mv^2/r

The Attempt at a Solution


Fc = mv^2/r

How can I solve this equation if I am missing m and v?

Write the 2 equations, one at the surface of the Earth where you know the value of g, and the other at altitude. You should be able to cancel a lot of things out... :-)
 
  • #4
MattHorbacz said:
The only force acting upon the object in this example is the force of Earth's gravity. You only need to use the equation F=GMm/r^2...remember F=ma, which means that acceleration=G*M/r^2 and mass =m. F=m*a=m*(GM/r^2)
and mass =m. F=m*a=m*(GM/r^2)[/QUOTE]

I don't understand this last part. Are you saying
m= m
F = ma = m * (GMr^2)

where did you get the mass from? I thought the equation for acceleration is G*M/r^2
 
  • #5
If the formula is a= GM/R^2

then i only need the value of M
 
  • #6
gcombina said:
If the formula is a= GM/R^2

then i only need the value of M
yes exactly...M in this case is the mass of the earth...it appears that they want you to look it up yourself...it happens to be 5.972*10^24 kg

ALSO! don't forget that your radius is actually radius of Earth + objects distance from earth
 
  • #7
If you have the acceleration due to gravity at one radius (like the surface of the Earth), then you can find the acceleration due to gravity at another radius using ratios. No need to know the actual mass of the Earth or the value of G.

Write the expression for the acceleration due to gravity for both locations using appropriate variables (so R1 and g for one, R2 and a for the other). Then make a ratio of the equations. All the common variables will cancel.
 

Related to Circular Orbit / Practice Quiz #4, Question 27

1. What is a circular orbit?

A circular orbit is the path that an object follows around a larger object, such as a planet or star, where the distance between the two remains constant. It is a type of orbit that has a perfectly circular shape, as opposed to an elliptical orbit which is more oval-shaped.

2. How is a circular orbit different from an elliptical orbit?

A circular orbit is a type of orbit with a constant distance between the two objects, while an elliptical orbit has a varying distance between the two objects. Additionally, a circular orbit has a perfectly circular shape, while an elliptical orbit is more oval-shaped.

3. What factors determine the size of a circular orbit?

The size of a circular orbit is determined by the mass of the larger object and the speed of the smaller object. The larger the mass of the object, the larger the orbit will be, and the faster the object is moving, the smaller the orbit will be.

4. Can objects in a circular orbit ever collide?

No, objects in a circular orbit cannot collide unless they are affected by outside forces, such as gravitational forces from other objects. This is because their orbits are stable and their paths do not cross.

5. How is a circular orbit beneficial for satellites and spacecraft?

A circular orbit is beneficial for satellites and spacecraft because it allows them to maintain a constant distance from the Earth, which is necessary for communication and data collection. It also requires less energy to maintain a circular orbit compared to an elliptical orbit.

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