Gravitational acceleration is related to distance via the equation:

 is the gravitational constant,  is the mass of the attracting object, and  is the distance from its center.

In this case, the initial distance (on the surface) is from the center of planet X, and the distance of the satellite is  from the center. We know that gravitational acceleration is proportional to the distance squared, and we know the acceleration at the surface. Using these values, we can solve for the acceleration on the satellite.

By expanding the equation for the acceleration on the satellite, we can see that it is equal to one-sixteenth the acceleration at the surface, based on our original equation. Substitute the value of the surface acceleration to get the final answer.

To solve this question, we will need to use the equation for acceleration:

In this case, the initial velocity will be equal to the final velocity, but opposite in magnitude. The initial velocity is in the upward direction, while the final velocity is downward.

Plug this value into the equation for acceleration.

The velocity value is constant, regardless of the planet. Substitute the acceleration for each planet to determine the change in the time variable.

We can see that the time of flight on the moon is equal to six times the time of flight on Earth.

The force due to gravity on any object can be given by the equation below.

 is the gravitational constant,  is the mass of the earth,  is the mass of the object, and  is the distance between the center of each object.

In our question, the only value to change is the radius of the new planet; both masses and  remain constant. The effect of doubling the radius on the force is given below.

The person's weight on the new planet would be one-fourth their weight on Earth.

To relate force and mass, we use Newton's second law:

We are given his weight (force) on the moon, and we are told the relative gravitational acceleration on the moon. Using these values, we can find the astronaut's mass.

Since mass is the same regardless of gravitational acceleration, this is the same as the astronaut's mass on the earth.

It may be helpful to start with a free-body diagram showing the forces acting on the person. We have the gravitational force, F_g, downwards, and the normal force of the scale, F_n, upwards.

Use these to write the net force equation.

First we need to solve for the person’s mass, m. When the elevator is moving at a constant rate, we are given that F_n = 500N, and we can solve for m.

.

When the elevator comes to rest, then we have an acceleration of -0.5 m/s² (downwards acceleration). Plugging this in, we can solve for the new F_n.

First, we need to determine what the apparent weight actually is. If we draw a free body diagram of the weights, we note that the force due to gravity acts downward and the force of the platform on the weights (the normal force) acts upward.

The normal force, F_N in the above diagram, is what a scale reads, and is thus the apparent weight. Finding the normal force is not as easy as equating it to the weight because the system is in free-fall, and thus accelerating. We can use Newton’s second law and equate it to acceleration to find the normal force.

Assuming downward is the positive y direction, we can solve for F_N.

mg – F_N = ma

F_N = mg – ma = 0 N

Logically, this makes sense. If an object is in free fall, the acceleration of the system is the same as the acceleration due to gravity. 

The acceleration of gravity is given by the equation , where G is constant.

For Earth, .

For the new planet, 

.

So, the acceleration is the same in both cases.

The force between two objects is calculated by:

Use our given variables:

The best way to think of a system in equilibrium is that it is displaying a constant velocity in both the horizontal and vertical directions. Both the skydiver and the tightrope walker have a constant velocity. The driver, however, is driving around a circular track. You must remember that velocity is a vector of magnitude and direction. Because the direction of the driver is changing, he is not driving at a constant velocity; thus, he is not in equilibrium, and is experiencing an acceleration (centripetal acceleration).

Keep in mind that the skydiver is experiencing both the downward force of gravity, and the upward force of air resistence on his deployed parachute. These forces cancel, producing a net force of zero, and allowing him to fall at a constant velocity in equilibrium.

Since the only acceleration on the object is to the west, we can conclude that there is no vertical acceleration on the object. As a result, the northern force must be offset by the third force. The only way to offset the northern force is to have a force pointing south, in any direction. As such, the third force could be directly south, southwest, or southeast as long as the southern component can counteract the northern force.

The third force would only need to be equal to the northern force if it was pointing directly due south, and we can only confirm that the third force is not completely offsetting the western force. As a result, we can only conclude that the third force has some magnitude in the southern direction.