Final answer:
To find the eccentricity of the orbit, we subtract the closest distance from the furthest and divide by the sum of both distances. The calculated eccentricity is approximately 0.01947, which shows a nearly circular orbit.
Step-by-step explanation:
The question is asking to calculate the eccentricity of a planet's orbit given its closest and furthest distances from the sun.
The eccentricity of an orbit is a measure of how much it deviates from being a perfect circle, and is defined as:
e = (d_{far} - d_{close}) / (d_{far} + d_{close})
Where d_{far} is the furthest distance and d_{close} is the closest distance from the sun.
Given the furthest point is 780,100,000 km and the closest point is 750,300,000 km, we can substitute these values into the formula:
e = (780,100,000 km - 750,300,000 km) / (780,100,000 km + 750,300,000 km)
e = 29,800,000 km / 1,530,400,000 km
e = 0.01947
Therefore, the eccentricity of the planet's orbit is approximately 0.01947. This indicates a nearly circular orbit, as the eccentricity is much closer to 0 than to 1.
Final answer in 20 words: The planet's orbit has an eccentricity of approximately 0.01947, indicating a nearly circular path around the sun.