Net Gravitational Force Problem

In summary, the first problem involves finding the net gravitational force exerted on a 60 kg object placed between a 200 kg object and a 700 kg object, as well as determining the position where the net force on the 60 kg object would be zero. The second problem involves finding the total force on a spacecraft near a black hole and the difference in gravitational fields acting on occupants at the front and back of the spacecraft. The equations used are the gravitational force equation and the equation for gravitational field strength.
  • #1
GreenLantern674
27
0
I have two problems: can anyone help?

First Problem:
A 200 kg object and a 700 kg object are separated by 0.700 m.
(a) Find the net gravitational force exerted by these objects on a 60.00 kg object placed midway between them.
(b) At what position (other than an infinitely remote one) can the 60.00 kg object be placed so as to experience a net force of zero?
m from the 700 kg object (on a line connecting the 200 kg and 700 kg objects)

I already solved for the first part. I got 1.63e-5 N, which was correct.
For the second part I thought you could just set the gravitational force equation, F=G(m1)(m2)/r^2 equal to zero, plug in my variables and solve for r, but I can't solve for r with that setup. Can anyone help?

Second Problem:
A spacecraft in the shape of a long cylinder has a length of 100 m and its mass with occupants is 1840 kg. It has strayed too close to a 1.0 m radius black hole having a mass 106 times that of the Sun (Fig. P11.8). The nose of the spacecraft points toward the black hole, and the distance between the nose and the black hole is 10.0 km.
(a) Determine the total force on the spacecraft .
(b) What is the difference in the gravitational fields acting on the occupants in the nose of the ship and on those in the rear of the ship, farthest from the black hole?

Again with this one I got the first answer, which was 2.55e17 N. For the second part I tried finding the gravitational force equation at the front end and then at the back end of the rocket and solving for the difference, but it didn't work. Also, it says the answer should be in Newtons per kilogram. What does that mean?
 
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  • #2
GreenLantern674 said:
First Problem:
A 200 kg object and a 700 kg object are separated by 0.700 m.
(a) Find the net gravitational force exerted by these objects on a 60.00 kg object placed midway between them.
(b) At what position (other than an infinitely remote one) can the 60.00 kg object be placed so as to experience a net force of zero?
m from the 700 kg object (on a line connecting the 200 kg and 700 kg objects)

I already solved for the first part. I got 1.63e-5 N, which was correct.
For the second part I thought you could just set the gravitational force equation, F=G(m1)(m2)/r^2 equal to zero, plug in my variables and solve for r, but I can't solve for r with that setup. Can anyone help?
That equation gives you the force between any two masses. You need to set the force due to the 700 kg object equal to the force due to the 200 kg object. You'll need to use that equation twice.

Second Problem:
A spacecraft in the shape of a long cylinder has a length of 100 m and its mass with occupants is 1840 kg. It has strayed too close to a 1.0 m radius black hole having a mass 106 times that of the Sun (Fig. P11.8). The nose of the spacecraft points toward the black hole, and the distance between the nose and the black hole is 10.0 km.
(a) Determine the total force on the spacecraft .
(b) What is the difference in the gravitational fields acting on the occupants in the nose of the ship and on those in the rear of the ship, farthest from the black hole?

Again with this one I got the first answer, which was 2.55e17 N. For the second part I tried finding the gravitational force equation at the front end and then at the back end of the rocket and solving for the difference, but it didn't work. Also, it says the answer should be in Newtons per kilogram. What does that mean?
They don't want the force on a particular mass, which would be measured in Newtons. They want the field strength, which is force per unit mass. (Just like the field strength of Earth's gravity near the surface is 9.8 m/s^2 = 9.8 N/kg.)
 
  • #3
Problem 1:
I tried setting them equal, but then you get (G*700*60)/r^2 = (G*200*60)/r^2. You can't solve for r since they cancel each other out. Am I doing something wrong?

Problem 2:
How do you solve for field strength? Is it just the force divided by the mass? If 9.8 m/s^2 equals 9.8 N/kg, then is the acceleration always equal to the field strength?
 
  • #4
GreenLantern674 said:
Problem 1:
I tried setting them equal, but then you get (G*700*60)/r^2 = (G*200*60)/r^2. You can't solve for r since they cancel each other out. Am I doing something wrong?
Those are different distances--don't use the same letter for both. Use r_1 & r_2, perhaps. (Hint: What must r_1 + r_2 equal?)

Problem 2:
How do you solve for field strength? Is it just the force divided by the mass?
Yes. Since the force is:
[tex]F = \frac{G Mm}{r^2}[/tex]
What's F/m?

If 9.8 m/s^2 equals 9.8 N/kg, then is the acceleration always equal to the field strength?
Yes, the acceleration due to gravity (for a freely falling object) equals the gravitational field strength.
 
  • #5
I got the second problem so thanks for that, but the first one is still giving me trouble. I set the two equations equal to each other. The 60 kg and G both canceled out so when I cross multiplied I got 700kg(r2)^2 = 200kg(r1)^2. Then I used the equation r2=0.7 m -r1 and substituted that in for r2. I got a big quadratic but when I solved for the final answer I got 1.5 m, which is outside of the area in between the two masses. Did I do anything wrong?
 
  • #6
GreenLantern674 said:
I got a big quadratic but when I solved for the final answer I got 1.5 m, which is outside of the area in between the two masses. Did I do anything wrong?
Either you came up with the wrong quadratic or you solved it incorrectly.

Also: Generally a quadratic has two solutions. Often only one of those solutions makes physical sense.
 
Last edited:

1. What is Net Gravitational Force?

Net Gravitational Force is the overall gravitational force experienced by an object in a given system. It is the vector sum of all the individual gravitational forces acting on the object from other objects in the system.

2. How is Net Gravitational Force calculated?

Net Gravitational Force is calculated using Newton's Law of Universal Gravitation, which states that the force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

3. What factors affect Net Gravitational Force?

The factors that affect Net Gravitational Force are the masses of the objects involved and the distance between them. The larger the masses of the objects and the closer they are to each other, the greater the Net Gravitational Force will be.

4. How does Net Gravitational Force affect the motion of objects?

Net Gravitational Force affects the motion of objects by causing them to accelerate towards each other. This acceleration is directly proportional to the Net Gravitational Force and inversely proportional to the mass of the object.

5. How is Net Gravitational Force different from other forces?

Net Gravitational Force is different from other forces because it is a non-contact force that acts over a distance and is always attractive. Other forces, such as friction or tension, are contact forces and can be either attractive or repulsive.

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