题目大意是：质量1200kg的空间探测器在火星表面引力是G0=4.43×10³N,问你在距火星表面3400km和6800km处的引力是多少?（火星半径3400km）
a.由于引力与距离的平方成反比.所以G1/G0 = (3400²)/(6800²)=1/4
∴G1=G0/4 ≈1.1×10³N
b.G2/G0=(3400²)/(10200²)=1/9
∴G2=G0/9≈4.9×10²N
写得比较简略,有什么疑问欢迎继续追问~
感谢采纳^_^不会推算引力与距离的平方成反比所以G1/G0 = (3400²)/(6800²)=1/4啊TAT求解
英语翻译
英语翻译
The gravitational force that acts on a 1200 kg space probe on the surface of Mars is
4.43 × 103 N.The radius of Mars is 3400 km.Without using the mass of Mars,determine the
gravitational force that acts on the space probe when it is:
a 3400 km above the surface of Mars
b 6800 km above the surface of Mars.
The gravitational force that acts on a 1200 kg space probe on the surface of Mars is
4.43 × 103 N.The radius of Mars is 3400 km.Without using the mass of Mars,determine the
gravitational force that acts on the space probe when it is:
a 3400 km above the surface of Mars
b 6800 km above the surface of Mars.
英语人气：726 ℃时间：2020-05-21 15:47:30
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