Universal Gravitational constant

In summary, the conversation discusses the calculation of the maximum height reached by a projectile launched from the Earth's surface at a speed of 10.1 km/s. The conversation mentions the use of the universal gravitational constant, the radius and mass of the Earth, and the equations for potential and kinetic energy. One participant suggests using Newton's Second Law and setting up equations in the x and y directions to find the distance and velocity equations, while another suggests using the conservation of energy. Despite some disagreements on the method, it is ultimately concluded that the initial calculation of the maximum height is incorrect and further examination is needed.
  • #1
bearhug
79
0
At the Earth's surface a projectile is launched straight up at a speed of 10.1 km/s. To what height will it rise?
Universal gravitational constant = 6.673e-11 N m^2/kg^2
Radius of the Earth = 6.370e+6 m
Mass of the Earth = 5.980e+24 kg

I know to use the equation U= - (Gm1m2)/ r
but I haven't been getting the right answer
originally I took into consideration that U=1/2mv^2 and set the two equations equal to each other and solved for r which I took to be R+h.
R = radius of Earth and h being the rest of the distance starting from the Earth's surface.

I would appreciate any help on this it's driving me crazy! :uhh:
 
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  • #2
The gravitational constant here is useless, well..we don't really need it. We already know that g=9,8, with that you should be able to find it.
 
  • #3
QuantumKing said:
The gravitational constant here is useless, well..we don't really need it. We already know that g=9,8, with that you should be able to find it.

Absolutely incorrect. The acceleration due to gravity is only 9.8 m/s^2 on the surface of the Earth -- as you go up in altitude, the acceleration due to gravity becomes smaller. An initial velocity of 10 km/s is too fast to permit the approximation that g is everywhere constant.

- Warren
 
  • #4
To the OP,

Use the equation for the force felt by the projectile due to the force of gravity. Use that to find its acceleration. Use the acceleration, along with its initial altitude and velocity, to solve its kinematic equation for velocity = 0. That corresponds to its highest altitude: the point at which it stops going up and starts coming down.

- Warren
 
  • #5
bearhug said:
At the Earth's surface a projectile is launched straight up at a speed of 10.1 km/s. To what height will it rise?
Universal gravitational constant = 6.673e-11 N m^2/kg^2
Radius of the Earth = 6.370e+6 m
Mass of the Earth = 5.980e+24 kg

I know to use the equation U= - (Gm1m2)/ r
but I haven't been getting the right answer
originally I took into consideration that U=1/2mv^2 and set the two equations equal to each other and solved for r which I took to be R+h.
R = radius of Earth and h being the rest of the distance starting from the Earth's surface.

I would appreciate any help on this it's driving me crazy! :uhh:

EDIT: This was under the faulty assumption that the speed was 10.1 m/s.

How to answer your question:

1) Draw a free body diagram for the projectile (make it a square or a ball or something simple) and draw the forces on it.

2) Newton's Second law: Once you've drawn your free body diagrams, and labeled the forces on it, you can set up an equation in the x and y directions to represent the directional components of that force.

3) Setup your equations based on the diagram. There should on be a y component, since the ball is thrown straight up with no wind or air drag.

4) There's two different things you can do with the equations. You can either use calculus to find your distance and velocity equations, or if you're in an algebra class, look up the equations of kinetics or the equations of motion, and use what you know (like initial velocity and final velocity will have a relationship with acceleration), and since you want your final velocity to be 0 (at the top of it's throw, it comes to a stop for an instant before it switches direction), you can find the time it takes for initial velocity to go to 0 given the the way it's acclerated downward by gravity.
 
Last edited:
  • #6
chroot said:
Absolutely incorrect. The acceleration due to gravity is only 9.8 m/s^2 on the surface of the Earth -- as you go up in altitude, the acceleration due to gravity becomes smaller. An initial velocity of 10 km/s is too fast to permit the approximation that g is everywhere constant.

- Warren

oh my, I didn't realize it was going 10 KILOmeters per second.
 
  • #7
Actually, the OP's initial argument was the easiest: it uses the conservation of energy.

At the surface of the earth, it has kinetic energy 1/2 mv^2. In reaching its maximum altitude, it converts all of this kinetic energy to gravitational potential energy, (-GMm)/r.

Upon closer inspection, it really looks like your method for determining altitude is totally correct -- yet you say you're getting the wrong answer. What answer did you get, and what leads you to believe it's wrong?

- Warren
 
  • #8
I set 1/2mv^2=(-GMm)/(R+h)
1/2m(1.01e4m/s)^2= (-6.673e-11)(5.98e24)m/(6.730e6 + h)
I solved for h and got 1.45e6m

This homework question is on a computer program that tells me whether It's right or not. That's why it's so frustrating.
 
  • #9
bearhug said:
I set 1/2mv^2=(-GMm)/(R+h)
1/2m(1.01e4m/s)^2= (-6.673e-11)(5.98e24)m/(6.730e6 + h)
I solved for h and got 1.45e6m

This homework question is on a computer program that tells me whether It's right or not. That's why it's so frustrating.

Looks like you've got the right idea, but your final answer is incorrect. Perhaps you're just making some kind of arithmetic mistake?

- Warren
 

What is the Universal Gravitational Constant?

The Universal Gravitational Constant, denoted by G, is a physical constant in Newton's law of universal gravitation. It defines the strength of the gravitational force between two objects with mass.

What is the numerical value of G?

The numerical value of G is approximately 6.674 × 10^-11 m^3 kg^-1 s^-2. This value is derived from various experiments and observations, and it is considered to be one of the most accurately measured physical constants.

How is G used in calculations?

G is used in calculations to determine the force of gravity between two objects with mass. It is also used to calculate the acceleration due to gravity on Earth's surface, as well as in astronomical calculations such as the gravitational force between celestial bodies.

Is G a universal constant?

Yes, G is considered to be a universal constant because its value does not change regardless of the location or mass of the objects involved. However, its value may vary in different theories of gravity, such as Einstein's theory of general relativity.

Why is the value of G important?

The value of G is important because it helps us understand and predict the behavior of large-scale objects in space, such as planets, stars, and galaxies. It also plays a crucial role in fields such as astrophysics and engineering, where gravity is a significant factor to consider.

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