Help with a Universal Gravitational Constant

In summary: I understand the theory but I need to memorise equations and such. I struggle with that.In summary, the conversation is about a student struggling with a question on calculating the orbital period for a satellite in an Astronomy GCSE course. The student is stuck on finding the value of G, the Universal Gravitational Constant, and has questions about the formula and units involved. Another student offers helpful advice on how to solve the problem and suggests using ratio reasoning to simplify the calculation. The conversation concludes with the student expressing gratitude for the assistance and acknowledging the need to improve on understanding and memorizing fundamental equations.
  • #1
Atominate
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I am doing an Astronomy GCSE. I have to work out the orbital period for a satellite. I have got quite far with it, but I am really stuck. I have spent hours at it! This is the question:-
A space laboratory is in circular orbit around the Earth at a distance of 6000km from the Earth's centre; its period is 2hr 12min. An orbital telescope is to be placed into orbit at 12,000km. Calculate its expected period.
I am at the point where I need to square root 4(pi)²r³/GM = T
A few questions. Is r the radius of the Earth (from core to surface) or core to satellite? How do I calculate GM? I know that M is the mass of the Earth (5.972 × 10^24 kg) but how do I work out G (the Universal Gravitational Constant?). I really need help and it is very frustrating. Any help or advice would be very much appreciated. Thanks!
 
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  • #2
"r" is usually said to be the distance from the core of Earth to the satellite. So r should be something like:

##r = R + h##
Where R is the radius of Earth and h is the height of the satellite above Earth's surface.

I'm assuming you have to calculate G yourself, correct? You have sufficient information to find G with the information provided for the space laboratory.
 
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  • #4
Atominate said:
I am doing an Astronomy GCSE. I have to work out the orbital period for a satellite. I have got quite far with it, but I am really stuck. I have spent hours at it! This is the question:-
A space laboratory is in circular orbit around the Earth at a distance of 6000km from the Earth's centre; its period is 2hr 12min. An orbital telescope is to be placed into orbit at 12,000km. Calculate its expected period.
I am at the point where I need to square root 4(pi)²r³/GM = T
A few questions. Is r the radius of the Earth (from core to surface) or core to satellite? How do I calculate GM? I know that M is the mass of the Earth (5.972 × 10^24 kg) but how do I work out G (the Universal Gravitational Constant?). I really need help and it is very frustrating. Any help or advice would be very much appreciated. Thanks!
The universal gravitational constant G is one of those fundamental numbers which describes the universe. You don't calculate it; its value is determined experimentally.

For your satellite problem, you just look up the value of G, either in a book or online.

https://en.wikipedia.org/wiki/Gravitational_constant
 
  • #5
It is possible to solve your problem without knowing the value of G or the mass of the Earth.
 
  • #6
faradayscat said:
"r" is usually said to be the distance from the core of Earth to the satellite. So r should be something like:

##r = R + h##
Where R is the radius of Earth and h is the height of the satellite above Earth's surface.

I'm assuming you have to calculate G yourself, correct? You have sufficient information to find G with the information provided for the space laboratory.

Okay. That's very helpful. Thank you.
 
  • #7
I
SteamKing said:
The universal gravitational constant G is one of those fundamental numbers which describes the universe. You don't calculate it; its value is determined experimentally.

For your satellite problem, you just look up the value of G, either in a book or online.

https://en.wikipedia.org/wiki/Gravitational_constant

I looked it up online, but it looks like M, kg and s are required?
 
  • #8
TonyS said:
It is possible to solve your problem without knowing the value of G or the mass of the Earth.

Oh. Could you please elaborate?
 
  • #9
Atominate said:
Oh. Could you please elaborate?
The physics of the problem allows you to find an expression for the orbital period as a function of r, that involves your unknown constants.
You know the values of T and r for one particular case, namely the spacestation. The same relation between T and r also holds for the telescope, with the same bunch of constants, so you can eliminate the constants by dividing T(telescope) by T(spacestation).
(Sorry I could make this clearer if I'd learned to use latex)
 
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  • #10
Atominate said:
I looked it up online, but it looks like M, kg and s are required?

Well, let's take a look at your formula:

Atominate said:
A space laboratory is in circular orbit around the Earth at a distance of 6000km from the Earth's centre; its period is 2hr 12min. An orbital telescope is to be placed into orbit at 12,000km. Calculate its expected period.
I am at the point where I need to square root 4(pi)²r³/GM = T

G = 6.67408 × 10-11 m3 / kg-s2

You want to find T, which is measured in seconds - check.
You'll have to look up the mass of the Earth in kilograms - check.
You know the distance from the center of the Earth to the satellite, but you need this in meters instead of kilometers - check.

Looks like G has all the right units. :wink:

From a practical standpoint, TonyS' post gives you a good tip for finding the new orbital period.

OTOH, you need to study up on fundamental constants, like G, if you want to do well in your astronomy course.
 
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  • #11
Atominate said:
A space laboratory is in circular orbit around the Earth at a distance of 6000km from the Earth's centre; its period is 2hr 12min. An orbital telescope is to be placed into orbit at 12,000km. Calculate its expected period.

Use ratio reasoning. The radius of the orbit is doubled, so what happens to the period?
 
  • #12
Mister T said:
Use ratio reasoning. The radius of the orbit is doubled, so what happens to the period?
Is the period doubled? That would make everything a lot simpler! Thanks.
 
  • #13
SteamKing said:
Well, let's take a look at your formula:
G = 6.67408 × 10-11 m3 / kg-s2

You want to find T, which is measured in seconds - check.
You'll have to look up the mass of the Earth in kilograms - check.
You know the distance from the center of the Earth to the satellite, but you need this in meters instead of kilometers - check.

Thanks very much. That helps a lot!
P.S. I completely agree. I am lacking with the more mathematical part of course. I will do further learning!
 
  • #14
TonyS said:
(Sorry I could make this clearer if I'd learned to use latex)

It is very clear now! Thanks.
 
  • #15
Atominate said:
I am doing an Astronomy GCSE. I have to work out the orbital period for a satellite.
A space laboratory is in circular orbit around the Earth at a distance of 6000km from the Earth's centre; its period is 2hr 12min. An orbital telescope is to be placed into orbit at 12,000km. Calculate its expected period.
If you studied Astronomy, you must be familiar with Kepler's Laws of planetary motion. What does the third Law say?
http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html
 
  • #16
Atominate said:
Is the period doubled? That would make everything a lot simpler! Thanks.
The square of the period is proportional to the cube of the radius, as you indicated in your original post, aside from a typo, I think. This is one of Kepler's laws, by the way.
 
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  • #17
Kepler's third law says that T2/R3 = constant = 4π2 /GM. I was okay with that part, it was just that I was unsure about GM, but I'm okay now as everyone has helped a lot with their replies.
 
  • #18
Since 2 hours 12 minutes is 12.2 hours, we have

##\frac{T^2}{R^3}=\frac{(12.2 \ \mathrm{h})^2}{(6000 \ \mathrm{km})^3}##.

That is the constant of proportionality. So when ##R## doubles to 12 000 km, ##T## does not double to 24.4 hours.
 

1. What is the Universal Gravitational Constant?

The Universal Gravitational Constant, also known as the gravitational constant or Newton's constant, is a physical constant that is used to determine the strength of the gravitational force between two objects with mass. It is denoted by the letter G and has a value of approximately 6.674 x 10^-11 Nm^2/kg^2.

2. How is the Universal Gravitational Constant calculated?

The value of the Universal Gravitational Constant is determined through experiments and measurements. One of the most famous experiments used to calculate G was the Cavendish experiment, which involved measuring the gravitational force between two masses using a torsion balance. Other methods, such as studying the motion of planets and stars, have also been used to calculate G.

3. What is the significance of the Universal Gravitational Constant?

The Universal Gravitational Constant is a fundamental constant in physics and plays a crucial role in understanding the behavior of objects in the universe. It is used in various equations, such as Newton's Law of Universal Gravitation, to calculate the force of gravity between two objects. G also helps us understand the structure and evolution of the universe, as well as the motion of celestial bodies.

4. Does the value of the Universal Gravitational Constant vary?

The value of G is considered to be a constant, meaning it does not change in different situations or locations in the universe. However, there have been some discrepancies in measurements of G, leading to ongoing research and debates about its true value. Some theories, such as the varying speed of light theory, suggest that the value of G may change over time.

5. How does the Universal Gravitational Constant relate to other constants in physics?

G is one of the fundamental constants in physics, along with the speed of light, Planck's constant, and the elementary charge. These constants are used to define the basic properties of the universe and are essential in understanding the laws of nature. G also plays a role in other important equations, such as Einstein's theory of general relativity and the equation for escape velocity.

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