Pyth, 10 bytes

VT.dC"¼€

Try it online!

There are exactly 31,557,600 seconds in the Julian astronomical year., totalling 31,557,600,000 seconds in 1000 years. 2^32 is about a tenth of this, so we just wait for 3,155,760,000 seconds ten times. Note that is a blank codepoint in the TIO, not sure why it translates to this on SE

Here you can see that C"¼ is equal to 3,155,760,000

And here is an example that waits for only 22 seconds using a similar method

Pyth, 8 bytes

.dC"XúÃ

Try it online!

Alternatively, this one just uses C"XúÃ for 31,557,600,000. I thought it was more in line with the spirit of the challenge to have a 2^32 limit, though

Pyth, 10 bytes

VT.dC"¼€

Try it online!

There are exactly 31,557,600 seconds in the Julian astronomical year., totalling 31,557,600,000 seconds in 1000 years. 2^32 is about a tenth of this, so we just wait for 3,155,760,000 seconds ten times. Note that is a blank codepoint in the TIO, not sure why it translates to this on SE

Here you can see that C"¼ is equal to 3,155,760,000

And here is an example that waits for only 22 seconds using a similar method

Pyth, 8 bytes

.dC"XúÃ

Try it online!

Alternatively, this one just uses C"XúÃ for 31,557,600,000. I thought it was more in line with the spirit of the challenge to have a 2^32 limit, though

The average length of a sidereal year, however, is 365.256363004 days or around 31,558,149.763 seconds, giving us a total of ~31,558,149,764 seconds in 1000 years. In this case, the more accurate solution would be

This: VT.dC"¼ê@

This:.dC"Y&(1 byte longer than Julian solution)

Pyth, 10 bytes

VT.dC"¼€

Try it online!

There are exactly 31,557,600 seconds in the Julian astronomical year., totalling 31,557,600,000 seconds in 1000 years. 2^32 is about a tenth of this, so we just wait for 3,155,760,000 seconds ten times.

Here you can see that C"¼ is equal to 3,155,760,000

And here is an example that waits for only 22 seconds using a similar method

Pyth, 10 bytes

VT.dC"¼€

Try it online!

There are exactly 31,557,600 seconds in the Julian astronomical year., totalling 31,557,600,000 seconds in 1000 years. 2^32 is about a tenth of this, so we just wait for 3,155,760,000 seconds ten times. Note that is a blank codepoint in the TIO, not sure why it translates to this on SE

Here you can see that C"¼ is equal to 3,155,760,000

And here is an example that waits for only 22 seconds using a similar method

Pyth, 8 bytes

.dC"XúÃ

Try it online!

Alternatively, this one just uses C"XúÃ for 31,557,600,000. I thought it was more in line with the spirit of the challenge to have a 2^32 limit, though