from urllib.parse import urlparse


def extract_s3_path_from_url(url):
    """
    Extracts the S3 object path from an S3 URL or URI.
    
    This function parses S3 URLs/URIs and returns just the object path portion,
    removing the protocol (s3://), bucket name, and any leading slashes.
    
    Args:
        url (str): The full S3 URI (e.g., 's3://eodata/path/to/file.jp2')
    
    Returns:
        str: The S3 object path (without protocol, bucket name and leading slashes)
    """
    # If it's not an S3 URI, return it unchanged
    if not url.startswith('s3://'):
        return url
    
    # Parse the S3 URI
    parsed_url = urlparse(url)
    
    # Ensure this is an S3 URL
    if parsed_url.scheme != 's3':
        raise ValueError(f"URL {url} is not an S3 URL")
    
    # Extract the path without leading slashes
    object_path = parsed_url.path.lstrip('/')
    
    return object_path