Question

How far above the surface of the Earth does an object have to be in order for it to have the same weight as it would have on the surface of the Moon? (Ignore any effects from the Earth's gravity for the object on the Moon's surface or from the Moon's gravity for the object above the Earth.)

Short answer

The object must be approximately 5.73 million meters (5730 km) above Earth's surface.

Step-by-step solution

Understand the weight formula

The weight of an object is given by the formula: \( W = mg \), where \( m \) is the mass of the object and \( g \) is the gravitational acceleration. On Earth's surface, \( g = 9.8 \, \text{m/s}^2 \) and on the Moon's surface, \( g = 1.63 \, \text{m/s}^2 \). The problem requires us to find the distance from Earth's surface where the gravitational acceleration is equal to that on the Moon.

Gravitational acceleration at a height

The gravitational acceleration at a distance \( r \) from the center of a body is given by: \( g_r = \frac{G M}{r^2} \). Here, \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \) and \( M \) is the Earth's mass approximately \( 5.972 \times 10^{24} \, \text{kg} \). To find the height, let \( g_r = 1.63 \, \text{m/s}^2 \) (same as on the Moon).

Solve for the distance from Earth's center

Set the formula for gravitational acceleration equal to the Moon's surface gravity: \( \frac{G M}{r^2} = 1.63 \). Solve the equation for \( r \):\[ r = \sqrt{\frac{G M}{1.63}}\]. Substitute \( G = 6.674 \times 10^{-11} \) and \( M = 5.972 \times 10^{24} \):\[ r = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{1.63}} \approx 1.21 \times 10^7 \, \text{m} \].

Compute the height above Earth's surface

The Earth's radius \( R \) is approximately \( 6.371 \times 10^6 \, \text{m} \). The height \( h \) above Earth's surface is given by \( h = r - R \):\[ h = 1.21 \times 10^7 - 6.371 \times 10^6 \approx 5.73 \times 10^6 \, \text{m} \].

Key concepts

Weight Formula

In physics, when we talk about the weight of an object, we are referring to the force that gravity exerts on it. This can be calculated using the weight formula: \[ W = mg \]

• \( W \) represents the weight in newtons (N).
• \( m \) is the mass of the object in kilograms (kg).
• \( g \) is the gravitational acceleration in meters per second squared (m/s²).

Gravitational acceleration (\( g \)) varies depending on where you are. For instance, on Earth's surface, \( g \) is approximately 9.8 m/s², while on the Moon's surface, it's about 1.63 m/s².
When solving problems like determining how far above Earth an object would need to be to weigh the same as it does on the Moon, understanding \( g \) on different celestial bodies is crucial. This allows us to set the gravitational forces equal and find out how gravitational effects change with distance.

Mass of the Earth

The mass of the Earth plays a vital role in calculating gravitational forces. It's a huge number and is approximately \( 5.972 \times 10^{24} \, \text{kg} \). This mass is used in many formulas that involve gravitational calculations. Applying the mass of Earth helps us understand how much force it can exert over objects placed on its surface or at any distance from it.By combining Earth's mass with Newton's universal law of gravitation, we can determine how weight changes at various distances. This is key when we need to compare the weight of an object on different celestial bodies, like in this exercise where we treat Earth's gravity to equal that of the Moon's at a specific height.

Gravitational Constant

In physics, the gravitational constant (\( G \)) is a critical part of understanding how masses attract each other. It is a very small number and is denoted as:\[ G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \]This constant is a key part of the equation used to determine the gravitational force between two masses. In the equation \[ g_r = \frac{G M}{r^2} \], it's combined with the mass of the Earth to help calculate the gravitational force at different distances from Earth's center.
Gravitational constant helps us understand how universal gravitation binds everything together, not just on Earth, but throughout the cosmos.

Radius of the Earth

A planet's radius is important in determining the gravitational pull experienced by objects on or near it. For Earth, this radius is about \( 6.371 \times 10^6 \, \text{m} \).
The radius is utilized to compute heights above the Earth's surface. In our problem, after finding the distance \( r \) from the Earth's center where gravitational acceleration equals that of the Moon's surface, we subtract Earth's radius to find the actual height above Earth's surface.Understanding Earth's size helps us explore gravitational effects at various distances, which is essential in many engineering and scientific fields, including satellite deployment and space exploration. This vital measurement allows us to bridge theoretical physics with practical, real-world applications.