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With what minimum velocity must a rocket be launched from the surface of the moon in order to not fall back down due to the moon's gravity?
The mass of the moon is and its radius is
.
Relevant equations:
For the rocket to escape the moon's gravity, its minimum total energy is zero. If the total energy is zero, the rocket will have zero final velocity when it is infinitely far from the moon. If total energy is less than zero, the rocket will fall back to the moon's surface. If total energy is greater than zero, the rocket will have some final velocity when it is infinitely far away.
For the minimum energy case as the rocket leaves the surface:
Rearrange energy equation to isolate the velocity term.
Substitute in the given values to solve for the velocity.
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What is the gravitational force of the sun on a book on the earth's surface if the sun's mass is
and the earth-sun distance is
?
Relevant equations:
Use the given values to solve for the force.
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Two spheres of equal mass are isolated in space, and are separated by a distance . If that distance is doubled, by what factor does the gravitational force between the two spheres change?
Newton's law of universal gravitation states:
We can write two equations for the gravity experienced before and after the doubling:
The equation for gravity after the doubling can be simplified:
Because the masses of the spheres remain the same, as does the universal gravitation constant, we can substitute the definition of Fg1 into that equation:
The the gravitational force decreases by a factor of 4 when the distance between the two spheres is doubled.
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Two spheres of equal mass are isolated in space. If the mass of one sphere is doubled, by what factor does the gravitational force experienced by the two spheres change?
Newton's law of universal gravitation states that:
We can write two equations representing the force of gravity before and after the doubling of the mass:
The problem gives us and we can assume that all other variables stay constant.
Substituting these defintions into the second equation:
This equation simplifies to:
Substituting the definition of Fg1, we see:
Thus the gravitational forces doubles when the mass of one object doubles.
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You are riding in an elevator that is accelerating upwards at , when you note that a block suspended vertically from a spring scale gives a reading of
.
What does the spring scale read when the elevator is descending at constant speed?
When the elevator accelerates upward, we know that an object would appear heavier. The normal force is the sum of all the forces added up, and in this case it is . We know that the normal force has two components, a component from gravity, and a component from the acceleration of the elevator. Using this equation, we can determine the mass of the block, which doesn't change:
is acceleration due to gravity and
is the acceleration of the elevator.
When substituting in the values, we get
Solving for , we get
Since the elevator is descending at constant speed, no additional force is applied, therefore the force that the spring scale reads is only due to gravity, which is calculated by:
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The mass and radius of a planet’s moon are and
respectively.
With what minimum speed would a bullet have to be fired horizontally near the surface of this moon in order for it to never hit the ground?
(Note: You can treat the moon as a smooth sphere, and assume there’s no atmosphere.)
To do this problem we have to realize that the force of gravity acting on the bullet is equal to the centripetal force. The equations for gravitational force and centripetal force are as follows:
If we set the two equations equal to each other, the small (mass of the bullet) will cancel out and
will disappear from the right side of the equation.
is the universal gravitational constant
is given to be
and
to be
.
If we plug everything in, we get
or
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The force on an object due to gravity on the moon is one-sixth of that found on Earth. What is the acceleration due to gravity on the moon?
We can use Newton's second law:
Set up equations for the force on the moon and the force on Earth:
Now we can use substitution:
From this, we can see that . Using the acceleration due to gravity on Earth, we can find the acceleration due to gravity on the moon.
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A large planet exerts a gravitational force five times stronger than that experienced on the surface of Earth. What is the weight of a 50kg object on this planet?
The weight of the object on Earth's surface is:
The force on the new planet is five times that on Earth, so we can simply multiply:
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A space woman finds herself in an unkown planet with gravity . If her weight on Earth is 500N, what is her weight on the unkown planet?
We know that the weight of an object is given by:
is the mass of the object and
is the gravitational acceleration of whatever planet the object happens to be on.
We know the gravity on the unkown planet, so the weight of the woman is given by:
We need only to find the mass of the woman to solve the problem. Since the mass of the woman is constant, we can use the information about her weight on Earth to figure out her mass.
Use this mass to solve for her weight on the new planet.
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A block with a mass of is traveling at
when it impacts the ground. From how many meters off the ground was the block dropped?
Round to the nearest whole number.
Set the gravitational potential energy and kinetic energy equal to each other and solve for the height.
Mass cancels.
Isolate the height and plug in our values.
Rounding this gives .
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A spherical asteroid has a hole drilled through the center as diagrammed below:
Refer to the diagram above. An object that is much smaller than the asteroid is released from rest at the surface of the asteroid, at point a. How do the velocity and acceleration of the object compare at point b at the surface, and point a, located at the center of the asteroid?
Because the gravitational force depends only on the mass beneath the object (which is the gravitational version of Gauss's Law for charge), the acceleration steadily decreases as the object falls, and drops to zero at the center. Nevertheless, the velocity keeps increasing as the object falls, it just does so more slowly.
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An object of mass is dropped from a tower. The object's drag force is given by
where
is a positive constant. What will the objects terminal velocity be?
To find terminal velocity, set the magnitude of the drag force equal to the magnitude of the force of gravity since when these forces are equal and opposite, the object will stop accelerating:
Solve for
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A ship of mass and an initial velocity of
is coasting to a stop. The water exerts a drag force on the ship. The drag force is proportional to the velocity:
where the negative sign indicates that the drag force acts in a direction opposite the motion. After the ship has coasted for a time equal to
, how fast (in terms of
) will the ship be moving?
The ship's equation of velocity (found by solving the first-order differential equation) is
Substitute and solve.
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If is the escape velocity from the surface of a planet of mass
and radius
. What is the velocity necessary for an object launched from the surface of a planet of the same mass, but with a radius that is
, to escape the gravitational pull of this smaller, denser planet?
Escape velocity is the velocity at which the kinetic energy of an object in motion is equal in magnitude to the gravitational potential energy. This allows us to set two equations equal to each other: the equation for kinetic energy of an object in motion and the equation for gravitational potential energy. Therefore:
In this case, . Substitute.
Since the radius of the new planet is or
the radius of the original planet, and they have the same masses, we can equate:
Where is the escape velocity of the larger planet and
is the escape velocity of the smaller planet.
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A comet is in an elliptical orbit about the Sun, as diagrammed below:
The mass of the comet is very small compared to the mass of the Sun.
In the diagram above, how does the net torque on the comet due to the Sun's gravitational force compare at the two marked points?
Since the direction of the gravitational force is directly towards the center of the Sun, it lies in the plane of the comet's orbit. Since torque is , the torque is always zero. That is why the comet's angular momentum is conserved in orbit.
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A satellite of mass is in an elliptical orbit about the Earth. It's velocity at perigee is
and its orbital radius at perigee is
. If the radius at apogee is
, what is its velocity at apogee?
Since the gravitational force cannot exert torque on the satellite about the Earth's center, angular momentum is conserved in this orbit:
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A satellite is in an elliptical orbit about the Earth. If the satellite needs to enter a circular orbit at the apogee of the ellipse, in what direction will it need to accelerate?
Since the speed of the satellite at apogee is too low for a circular orbit at that orbital radius, the satellite needs to speed up to circularize the orbit.
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A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?
When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.
In addition, the rocket experiences the downward force of its own weight. This is given by:
We know that .
We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.
We know that (since
is directed upwards and
is directed downwards).
Finally we solve for acceleration:
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Three boxes tied by two ropes move across a frictionless surface pulled by a force as shown in the diagram. Which of the following is an expression of the acceleration of the system?
Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.
A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between and
as
. Similarly, we will label the tension experienced by the rope between
and
as
.
For the box on the right (of mass ) we know that the force
pulls to the right and the rope pulls it to the left with a tension of
.
We get the equation: .
is the mass of the box and
is its acceleration (which is the same for the other boxes).
For the middle box (of mass ) we have a rope pulling on it on the right with tension
and a rope pulling on the left with tension
.
So we get the equation: .
is the mass of the box and
is its acceleration (which is the same for the other boxes).
Finally, for the box on the left (of mass ) we have only one rope pulling it to the right with a tension
.
So we get the equation: .
is the mass of the box and
is its acceleration (which is the same for the other boxes).
Now it is just a matter of simple substitution. We have three equations:
From them we can get an expression for the force. Isolate in the second equation.
Substitute the expression of from the first equation and simplify.
Use this value of in the third equation to get an expression for the force.
So we have:
Solve for acceleration.
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A box is being pushed against a wall by a force as shown in the picture. If the coefficient of static friction between the box and the wall is
, which of the following expressions represents what the force must be for the box not to fall?
This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force . The box experiences the downward force of its own weight given by
. Finally, since the box is trying to slide down, friction
opposes this motion. The following diagram shows these forces:
The key is that the box should NOT move.
For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.
Now, on the vertical axis we have fiction and the object's weight:
Finally, we use the equation for friction force to solve the problem.
Substitute the weight, since it is equal to the force of friction.
Isolate the normal force.
Since the normal force is equal to the applied force, this is our final expression.
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