Problem 32 - College Physics

Question

The asteroid 243 Ida has a mass of about \(4.0 \times 10^{16} \mathrm{~kg}\) and an average radius of about \(16 \mathrm{~km}\) (it's not spherical, but you can assume it is). (a) Calculate the acceleration of gravity on 243 Ida. (b) What would an astronaut whose earth weight is \(650 \mathrm{~N}\) weigh on 243 Ida? (c) If you dropped a rock from a height of \(1.0 \mathrm{~m}\) on 243 Ida, how long would it take for the rock to reach the ground? (d) If you can jump \(60 \mathrm{~cm}\) straight up on earth, how high could you jump on 243 Ida? (Assume the asteroid's gravity doesn't weaken significantly over the distance of your jump.)

Step-by-Step Solution

Answer

The gravitational acceleration on Ida is approximately 0.279 m/s². An astronaut would weigh 18.49 N on Ida. It takes about 2.68 seconds for a rock to fall 1 meter. You could jump approximately 21.1 meters high on Ida.

Step 1: Calculate the gravitational acceleration on 243 Ida

The formula for gravitational acceleration on a spherical body is given by:\[ g = \frac{G \cdot M}{r^2} \]where:- \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} \)- \( M \) is the mass of the body, \( 4.0 \times 10^{16} \mathrm{~kg} \)- \( r \) is the radius of the body, \( 16000 \mathrm{~m} \)Substituting the values:\[ g = \frac{6.674 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{~s}^{-2} \times 4.0 \times 10^{16} \mathrm{~kg}}{(16000 \mathrm{~m})^2} \approx 0.279 \mathrm{~m/s^2} \]

Step 2: Calculate the astronaut's weight on 243 Ida

Weight on a body is the product of mass and the local gravitational acceleration:\[ W = m \cdot g \]First, calculate the mass of the astronaut using Earth's gravity (\( g_{\text{earth}} \approx 9.81 \mathrm{~m/s^2} \)):\[ m = \frac{650 \mathrm{~N}}{9.81 \mathrm{~m/s^2}} \approx 66.29 \mathrm{~kg} \]Now calculate the weight on 243 Ida using its gravity \( g = 0.279 \mathrm{~m/s^2} \):\[ W_{\text{Ida}} = 66.29 \mathrm{~kg} \times 0.279 \mathrm{~m/s^2} \approx 18.49 \mathrm{~N} \]

Step 3: Calculate the time for the rock to fall 1 meter

To calculate the time taken for a rock to fall from a height using gravitational acceleration, we use the equation:\[ s = \frac{1}{2} g t^2 \]where:- \( s = 1.0 \mathrm{~m} \)Solving for \( t \) gives:\[ t = \sqrt{\frac{2s}{g}} = \sqrt{\frac{2 \times 1.0 \mathrm{~m}}{0.279 \mathrm{~m/s^2}}} \approx 2.68 \mathrm{~s} \]

Step 4: Calculate the jump height on 243 Ida

To determine how high you could jump on 243 Ida, first calculate kinetic energy equivalence. The potential energy on Earth at max jump height is converted to kinetic energy:\[ mgh_{\text{earth}} = mgh_{\text{Ida}} \]where \( h_{\text{earth}} = 0.6 \mathrm{~m} \). Solving for \( h_{\text{Ida}} \):\[ h_{\text{Ida}} = \frac{g_{\text{earth}}}{g_{\text{Ida}}} \times h_{\text{earth}} = \frac{9.81 \mathrm{~m/s^2}}{0.279 \mathrm{~m/s^2}} \times 0.6 \mathrm{~m} \approx 21.1 \mathrm{~m} \]

Key Concepts

Asteroidal Gravity

Asteroidal gravity refers to the gravitational pull exerted by an asteroid. Although asteroids are much smaller than planets and have far less mass, they still have a gravitational field. Gravity depends on two factors: mass and radius. The acceleration of gravity ( g ) on an asteroid can be calculated using the formula:\[g = \frac{G \cdot M}{r^2}\]where G is the gravitational constant, M is the mass of the asteroid, and r is its radius. For instance, the asteroid 243 Ida, despite its small size compared to Earth, has its own gravitational pull, calculated to be approximately 0.279 m/s². This value is much smaller than Earth's gravitational acceleration, which is about 9.81 m/s². However, Ida's gravity is significant enough to affect objects on its surface.

Weight Calculation

Weight on another celestial body is the force exerted by gravity on an object. This is different from mass, which remains constant regardless of location. When calculating how much someone would weigh on an asteroid like 243 Ida, you must first determine their mass based on their weight on Earth.

• First, calculate the mass m using Earth gravity: \[m = \frac{650 \mathrm{~N}}{9.81 \mathrm{~m/s^2}} \approx 66.29 \mathrm{~kg}\]
• Then, use the smaller gravitational force on the asteroid to find the new weight W : \[W_{\text{Ida}} = m \cdot g_{\text{Ida}} \approx 66.29 \mathrm{~kg} \times 0.279 \mathrm{~m/s^2} \approx 18.49 \mathrm{~N}\]

Thus, an astronaut weighing 650 N on Earth would weigh only 18.49 N on 243 Ida due to its much weaker gravity.

Free Fall Time

Free fall time refers to how long it takes for an object to reach the surface when dropped from a certain height. On the asteroid 243 Ida, we can calculate this using the equation for uniform acceleration: \[s = \frac{1}{2} g t^2\]This formula relates the distance fallen s , the gravitational acceleration g , and the time t . To find how long it takes a rock to fall from a 1-meter height on Ida:

• Rearrange to solve for t :\[t = \sqrt{\frac{2s}{g}}\]
• Plug in the given values: \[t = \sqrt{\frac{2 \times 1.0 \mathrm{~m}}{0.279 \mathrm{~m/s^2}}} \approx 2.68 \mathrm{~s}\]

This shows that it takes approximately 2.68 seconds for an object to fall 1 meter on 243 Ida, much longer compared to Earth, due to the asteroid’s weaker gravity.

Jump Height Comparison

Jump height comparison examines how differently gravity affects how high you can jump on various celestial bodies. Given Ida’s weaker gravity compared to Earth, you can actually jump higher. This is calculated by equating the energies involved in jumping:

• On Earth, the energy used to jump a certain height is converted into gravitational potential energy: \[mgh_{\text{earth}} = mgh_{\text{Ida}}\]
• Rearrange to find h_{\text{Ida} :\[h_{\text{Ida}} = \frac{g_{\text{earth}}}{g_{\text{Ida}}} \times h_{\text{earth}}\]
• Calculating gives:\[h_{\text{Ida}} = \frac{9.81 \mathrm{~m/s^2}}{0.279 \mathrm{~m/s^2}} \times 0.6 \mathrm{~m} \approx 21.1 \mathrm{~m}\]

Therefore, if you can jump 0.6 meters on Earth, you could leap to a height of about 21.1 meters on 243 Ida!