# Gravitation problem about throwing an object upward

• xnitexlitex
In summary, the problem involves an astronaut throwing a rock on the surface of a small spherical moon. They want to know the maximum height the rock will reach. The relevant equations are g=GM/r2 and conservation of energy. Using the kinematic equation did not work due to the changing acceleration caused by the small size of the moon.
xnitexlitex

## Homework Statement

Standing on the surface of a small spherical moon whose radius is 6.00*104 m and whose mass is 7.50*1018 kg, an astronaut throws a rock of mass 2.05 kg straight upward with an initial speed 38.5 m/s. (This moon is too small to have an atmosphere.) What maximum height above the surface of the moon will the rock reach?

g=GM/r2

## The Attempt at a Solution

I calculated g, then used the kinematic equation Δy = (vf2 - vi2) / 2a, but it didn't work! Did I use the wrong equations?

xnitexlitex said:

## Homework Statement

Standing on the surface of a small spherical moon whose radius is 6.00*104 m and whose mass is 7.50*1018 kg, an astronaut throws a rock of mass 2.05 kg straight upward with an initial speed 38.5 m/s. (This moon is too small to have an atmosphere.) What maximum height above the surface of the moon will the rock reach?

g=GM/r2

## The Attempt at a Solution

I calculated g, then used the kinematic equation Δy = (vf2 - vi2) / 2a, but it didn't work! Did I use the wrong equations?

Yup. The kinematic equations involving g that we use near the surface of the Earth are valid when g remains essentially constant over the trajectory of the projectile. This holds because the relative change in the radial distance from the center of the Earth is negligible, and g = G*M/r.

For your moon, the radius is relatively small and the upward launch speed will take the projectile a distance that is not negligible with respect to that radius.

You would be better to use a conservation of energy approach, using the appropriate expression for the gravitational potential energy as a function of radial distance.

## 1. What is the equation for calculating the height of an object thrown upward?

The equation for calculating the height of an object thrown upward is h = v2sin2(θ)/2g, where h is the height, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

## 2. How does the mass of the object affect its height when thrown upward?

The mass of the object does not affect its height when thrown upward. The height is determined by the initial velocity and the angle of projection, not the mass of the object.

## 3. What is the maximum height an object can reach when thrown upward?

The maximum height an object can reach when thrown upward is h = v2/2g, where h is the maximum height, v is the initial velocity, and g is the acceleration due to gravity. This occurs when the object reaches its highest point before falling back down.

## 4. How does air resistance affect the height of an object thrown upward?

Air resistance can reduce the height of an object thrown upward by slowing down its velocity. This is due to the force of air resistance acting in the opposite direction of the object's motion and reducing its overall kinetic energy.

## 5. What is the difference between throwing an object upward on Earth versus on the moon?

The main difference between throwing an object upward on Earth versus on the moon is the acceleration due to gravity. On Earth, the acceleration due to gravity is 9.8 m/s2, while on the moon it is only 1.6 m/s2. This means that an object thrown upward on the moon will reach a lower maximum height and take longer to fall back down compared to throwing it upward on Earth.

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