When the Hope probe was launched into its elliptical orbit around Mars it had an initial mass of 1350 kg. At its closest position, the distance of separation between the center of... When the Hope probe was launched into its elliptical orbit around Mars it had an initial mass of 1350 kg. At its closest position, the distance of separation between the center of Mars and the Hope probe is 20,000 km. Calculate the potential energy of the probe. (Take M_Mars = 6.42 × 10^23 kg.) When the Hope probe was launched into its elliptical orbit around Mars it had an initial mass of 1350 kg. At its furthest position, the distance of separation between the center of Mars and the Hope probe is 43,000 km. Calculate the potential energy of the probe. (Take M_Mars = 6.42 × 10^23 kg.) Read more

Steps to Solve

1. Understand the gravitational potential energy formula The formula for gravitational potential energy (U) is given by:

$$ U = -\frac{G \cdot M \cdot m}{r} $$

where:

• $U$ is the potential energy,
• $G$ is the gravitational constant, $6.674 \times 10^{-11} , \text{N m}^2/\text{kg}^2$,
• $M$ is the mass of the celestial body (Mars),
• $m$ is the mass of the object (Hope probe),
• $r$ is the distance from the center of the mass to the object.

2. Insert values for the closest distance (Question 1a) For the first part (1a):

• Mass of Mars, $M = 6.42 \times 10^{23} , \text{kg}$,
• Mass of the probe, $m = 1350 , \text{kg}$,
• Distance, $r = 20000 , \text{km} = 20000 \times 10^3 , \text{m}$.

Substituting these values into the formula gives:

$$ U = -\frac{6.674 \times 10^{-11} \cdot 6.42 \times 10^{23} \cdot 1350}{20000 \times 10^3} $$

3. Calculate and simplify the first part Now, performing the calculations step-by-step:

• Calculate the numerator:

$$ 6.674 \times 10^{-11} \cdot 6.42 \times 10^{23} \cdot 1350 $$

• Calculate the denominator:

$$ 20000 \times 10^3 $$

• Finally, calculate $U$.

4. Insert values for the furthest distance (Question 1b) For the second part (1b): Here, the distance, $r = 43000 , \text{km} = 43000 \times 10^3 , \text{m}$.

Substituting these into the formula gives:

$$ U = -\frac{6.674 \times 10^{-11} \cdot 6.42 \times 10^{23} \cdot 1350}{43000 \times 10^3} $$

5. Calculate and simplify the second part Again, perform the calculations for this second scenario:

• Using the same numerator as before.
• Calculate the new denominator:

$$ 43000 \times 10^3 $$

• Compute $U$ for this second distance.