At a Given Point Above Earth's Surface,the Free-Fall Acceleration Is $7.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$

question 105

Multiple Choice

At a given point above Earth's surface,the free-fall acceleration is equal to $7.8 \mathrm {~m} / \mathrm { s } ^ { 2 }$ What is the altitude of this point above Earth's surface? (G = 6.67 × 10^-11 N ∙ m²/kg²,M_earth = 5.97 × 10²⁴ kg,R_earth = 6.38 × 10⁶ m)

• A)770 km
• B)970 km
• C)1500 km
• D)2000 km
• E)2400 km

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