Question
In 1966, a special research cannon built in Arizona shot a projectile to a height of 180 km above Earth’s
surface. The potential energy associated with the projectile when its altitude was 10.0 percent of the maximum
height was 1.48 × 107J.What was the projectile’s mass? Assume that constant free-fall acceleration at this
altitude is the same as at sea level.
surface. The potential energy associated with the projectile when its altitude was 10.0 percent of the maximum
height was 1.48 × 107J.What was the projectile’s mass? Assume that constant free-fall acceleration at this
altitude is the same as at sea level.
Answers
GPT-4o mini
To find the mass of the projectile, we can use the formula for gravitational potential energy, which is given by:
\[
PE = mgh
\]
Where:
- \( PE \) is the potential energy,
- \( m \) is the mass of the object,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) at sea level, and we will assume it remains constant at 180 km),
- \( h \) is the height above the reference point (in this case, the height above the Earth's surface).
### Step 1: Calculate the height at 10.0 percent of the maximum height
The maximum height of the projectile, according to the problem, is \( 180 \, \text{km} \):
\[
h_{\text{max}} = 180 \, \text{km} = 180,000 \, \text{m}
\]
10 percent of the maximum height is:
\[
h = 0.10 \times h_{\text{max}} = 0.10 \times 180,000 \, \text{m} = 18,000 \, \text{m}
\]
### Step 2: Use the potential energy formula
We are given that the potential energy at this height is \( PE = 1.48 \times 10^7 \, \text{J} \).
Substituting the known values into the potential energy formula:
\[
1.48 \times 10^7 = m \cdot (9.81) \cdot (18,000)
\]
### Step 3: Solve for \( m \)
Now we will solve for \( m \):
\[
m = \frac{1.48 \times 10^7}{9.81 \times 18,000}
\]
Calculating the denominator:
\[
9.81 \times 18,000 = 176580
\]
Now substituting back into the equation for \( m \):
\[
m = \frac{1.48 \times 10^7}{176580} \approx 83.78 \, \text{kg}
\]
### Conclusion
The mass of the projectile is approximately:
\[
\boxed{83.8 \, \text{kg}}
\]
\[
PE = mgh
\]
Where:
- \( PE \) is the potential energy,
- \( m \) is the mass of the object,
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) at sea level, and we will assume it remains constant at 180 km),
- \( h \) is the height above the reference point (in this case, the height above the Earth's surface).
### Step 1: Calculate the height at 10.0 percent of the maximum height
The maximum height of the projectile, according to the problem, is \( 180 \, \text{km} \):
\[
h_{\text{max}} = 180 \, \text{km} = 180,000 \, \text{m}
\]
10 percent of the maximum height is:
\[
h = 0.10 \times h_{\text{max}} = 0.10 \times 180,000 \, \text{m} = 18,000 \, \text{m}
\]
### Step 2: Use the potential energy formula
We are given that the potential energy at this height is \( PE = 1.48 \times 10^7 \, \text{J} \).
Substituting the known values into the potential energy formula:
\[
1.48 \times 10^7 = m \cdot (9.81) \cdot (18,000)
\]
### Step 3: Solve for \( m \)
Now we will solve for \( m \):
\[
m = \frac{1.48 \times 10^7}{9.81 \times 18,000}
\]
Calculating the denominator:
\[
9.81 \times 18,000 = 176580
\]
Now substituting back into the equation for \( m \):
\[
m = \frac{1.48 \times 10^7}{176580} \approx 83.78 \, \text{kg}
\]
### Conclusion
The mass of the projectile is approximately:
\[
\boxed{83.8 \, \text{kg}}
\]