SpaceX pursues $40B in financing to buy Nvidia chips: Report
SpaceX aims to raise $40B, including $10B in loans and $30B in debt, to buy AI chips from Nvidia, according to a Financial Times report. Apollo Global Management will lead the financing, with Pimco among potential lenders. SpaceX shares fell 1% in extended trading, while Nvidia's stock rose 0.5%. SpaceX plans to use Nvidia hardware exclusively for its data centers.
How this was made

The 30-second read
Why it matters
The announcement underscores the massive capital requirements of the AI boom and validates Nvidia's position as the leading supplier of AI GPUs.
Market read
The financing deal signals sustained demand for AI hardware, likely supporting Nvidia's stock and the broader semiconductor sector.
What to watch
Potential regulatory scrutiny of large AI‑related financing and the execution risk of SpaceX's debt issuance.
Background
SpaceX is seeking up to $40 billion, led by Apollo Global Management, to fund a massive purchase of Nvidia AI chips for its data centers.
Ticker impact
Nvidia shares rose 0.5% in extended trading after the report that SpaceX plans a $40 billion financing to buy Nvidia AI chips.
slight upward pressure as the market prices in increased AI chip demand
The deal represents a multi‑billion dollar order from a high‑profile customer, reinforcing Nvidia's growth narrative.
Market effects
Boosts the AI‑chip sector outlook, reinforcing bullish bias on semiconductor stocks.
U.S. tech equities may see modest gains as the financing underscores AI infrastructure spending.
Highlights the scale of global AI capital needs, supporting worldwide demand for high‑performance GPUs.
Counterpoint
If the financing terms prove costly or the loan market tightens, the deal could strain Nvidia's pricing power.
Key entities
- companySpaceX
Rocket and spacecraft manufacturer planning a $40 billion financing to buy Nvidia chips.
- asset_managerApollo Global Management
Lead arranger for SpaceX's financing.
- companyNvidia
Supplier of AI GPUs that will benefit from the large order.




