Calculating Gravitational Field Constants for Mars, Jupiter, and Mercury

Click For Summary
To calculate the gravitational field constants for Mars, Jupiter, and Mercury, use the formula g = Gm/r^2, where G is the gravitational constant, m is the mass of the planet, and r is its radius. The gravitational field constants are approximately 3.61 m/s² for Mars, 24.6 m/s² for Jupiter, and 3.31 m/s² for Mercury. It is essential to apply Newton's Second Law of Motion alongside Newton's Universal Law of Gravitation. By substituting the mass of the planet into the equation, one can eliminate the second mass and simplify the calculation. Understanding this approach is crucial for accurately determining gravitational constants.
Dyna
Messages
9
Reaction score
0
1. Calculate the gravitational field constants for the following:
Mars (r=3.43 x 10^6m , m= 6.37 x 10^23 kg)
Jupiter (r= 7.18 x 10^7 , m = 1.90 x 10^27kg)
Mercury (r=2.57 x 10^6m , m= 3.28 x 10^23 kg)

2. F = mg
Fg = Gm1m2
r^2

3. So i tried subbing these given masses into the second equation but it does not give me the correct answer.
Am i not using the correct equation?
I only just learned this in school on Friday and it was quite confusing as to what i should be using to find my answers.
The answer key says:
Mars: 3.61 m/s^2
jupiter: 24.6m/s^2
mercury: 3.31m/s^2
 
Physics news on Phys.org
Use Newton's Second Law of Motion in conjunction with Newton's Universal Law of Gravitation.
 
In other words, since you're only given one mass (the planet's mass, call it m2), you've got to get rid of the second mass m1. You have the right equation to do this, F=mg. If you sub that into the left side, one of your m's will drop out, and you're left with g on the left side, and the correct answer on the right :)
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 23 ·
Replies
23
Views
4K
  • · Replies 6 ·
Replies
6
Views
16K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
2K
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
15
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K