The acceleration due to gravity at sea level is g = 9.81 m/s2. The radius of the earth is 6370 km. The universal gravitation constant is G = 6.67 × 10−11 N-m2/kg2. Use this information to determine the mass of the earth.
A person weighs 180 lb at sea level. The radius of the earth is 3960 mi. What force is exerted on the person by the gravitational attraction of the earth if he is in a space station in orbit 200 mi above the surface of the earth?
The Correct Answer and Explanation is :
Part 1: Mass of the Earth
We can use the formula for gravitational force to determine the mass of the Earth. The force due to gravity between two masses is given by:
[
F = \frac{G \cdot M \cdot m}{r^2}
]
Where:
- (F) is the gravitational force between two objects (in this case, the force experienced by an object due to Earth’s gravity).
- (G) is the universal gravitational constant: (6.67 \times 10^{-11} \, \text{N-m}^2/\text{kg}^2).
- (M) is the mass of the Earth (which we need to solve for).
- (m) is the mass of the object experiencing the force (in this case, a 1 kg object).
- (r) is the radius of the Earth: 6370 km or (6.37 \times 10^6 \, \text{m}).
The force on a 1 kg mass at the Earth’s surface is equal to the weight of the object, which is (9.81 \, \text{N}). We rearrange the formula to solve for the Earth’s mass:
[
M = \frac{F \cdot r^2}{G \cdot m}
]
Substituting the known values:
[
M = \frac{9.81 \, \text{N} \cdot (6.37 \times 10^6 \, \text{m})^2}{6.67 \times 10^{-11} \, \text{N-m}^2/\text{kg}^2 \cdot 1 \, \text{kg}}
]
[
M = \frac{9.81 \cdot 4.06 \times 10^{13}}{6.67 \times 10^{-11}}
]
[
M \approx 5.97 \times 10^{24} \, \text{kg}
]
Thus, the mass of the Earth is approximately (5.97 \times 10^{24} \, \text{kg}).
Part 2: Gravitational Force on the Person in Space
To calculate the force exerted on a person by Earth’s gravitational attraction in orbit, we use the same formula:
[
F = \frac{G \cdot M \cdot m}{r^2}
]
Where:
- (r) is the distance from the center of the Earth to the person. This is the sum of the Earth’s radius and the altitude of the space station: (r = 3960 \, \text{mi} + 200 \, \text{mi} = 4160 \, \text{mi}), which we convert to meters: (4160 \, \text{mi} = 6.69 \times 10^6 \, \text{m}).
- (M = 5.97 \times 10^{24} \, \text{kg}) (the mass of the Earth).
- (m = 180 \, \text{lb} = 81.65 \, \text{kg}) (the person’s mass converted to kilograms).
Substituting the known values into the equation:
[
F = \frac{6.67 \times 10^{-11} \cdot 5.97 \times 10^{24} \cdot 81.65}{(6.69 \times 10^6)^2}
]
First, calculate the denominator:
[
(6.69 \times 10^6)^2 = 4.47 \times 10^{13}
]
Now calculate the force:
[
F = \frac{6.67 \times 10^{-11} \cdot 5.97 \times 10^{24} \cdot 81.65}{4.47 \times 10^{13}} \approx 466.7 \, \text{N}
]
Thus, the gravitational force on the person in the space station is approximately 467 N.
Explanation
Gravitational force decreases with distance, so the force on the person in space is less than what they would experience on Earth’s surface. This calculation demonstrates how the force of gravity weakens as the person moves farther from the Earth’s center. Despite being in space, the person is still experiencing a significant gravitational pull, though less than at sea level.