# Object falling; acceleration based on x

• Sheldinoh
In summary, the conversation discusses the calculation of the speed at which a rock will impact the surface of the moon when released from a height of 3R above the surface. The formula used takes into account the acceleration due to gravity at a distance x from the center of the moon, with g0 being 1.63 m/s2 and R being 3200 km for the moon. The correct answer is found to be 2.797 km/s.

## Homework Statement

For a spherical celestial object of radius R, the acceleration due to gravity g at a distance x from the center of the object is g = (g0)(R^2)/(x^2), where g0 is the acceleration due to gravity at the object's surface and x > R. For the moon, take g0 = 1.63 m/s2 and R = 3200 km. If a rock is released from rest at a height of 3R above the lunar surface, with what speed does the rock impact the moon? Hint: Its acceleration is a function of position and increases as the object falls. So do not use constant acceleration free-fall equations, but go back to basics.

## Homework Equations

.5(v^2-v0^2) = (g0)(R^2)(1/x-1/x0)

## The Attempt at a Solution

i said that v0=0 and solved for v, x=3R x0=R, i got .899km/s. apparently that is incorrect.

What's the difference between height and radial distance?

2R, so would i plug 2R in as x and 0 for x0?

Sheldinoh said:
2R, so would i plug 2R in as x and 0 for x0?

What's the radial distance to the surface of the Moon?

What's the radial distance to an object at height 3R above the Moon's surface?

What's the radial distance to the surface of the Moon?

R

What's the radial distance to an object at height 3R above the Moon's surface?

4R

I see now, i think.
so x0=4R, x=R right?

i got 2.797km/s is that the correct answer?

It looks good.

Does the system care if you enter the result as km/sec or m/sec?

it wants the answer in km/s

Is it fussy about significant figures?

i got it, i think i plugged it into the calculator wrong.

thanks for all the help

## 1. What is the formula for calculating acceleration based on distance (x) for an object falling?

The formula for calculating acceleration based on distance (x) for an object falling is a = 2x/t^2, where a is the acceleration, x is the distance, and t is the time.

## 2. How does the acceleration of an object falling change as it travels a greater distance?

The acceleration of an object falling will remain constant (9.8 m/s^2) regardless of the distance it travels. This is due to the force of gravity acting on the object.

## 3. What factors can affect the acceleration of an object falling based on distance (x)?

The acceleration of an object falling based on distance (x) can be affected by the mass of the object, air resistance, and the gravitational pull of the Earth.

## 4. Can the acceleration of an object falling be negative?

Yes, the acceleration of an object falling can be negative if the object is moving in the opposite direction of the positive direction chosen for the calculation. This is often seen when an object is thrown upwards and begins to fall back down.

## 5. How is the acceleration of an object falling related to its velocity?

The acceleration of an object falling is directly related to its velocity. As the object falls, its velocity will increase due to the constant acceleration. This can be represented by the equation v = gt, where v is the velocity, g is the acceleration due to gravity, and t is the time.

• Introductory Physics Homework Help
Replies
10
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
23
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
4K
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
10
Views
148
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
19
Views
2K