The earth revolves round the sun with constant speed on an orbit of radius r=1.5x10^11m. Find the earth's acceleration towards the sun.
v=sTv = \frac{s}{T}v=Ts
s=2πrs = 2\pi rs=2πr
T=1 year ≈365∗24∗3600sT = 1 \text{ year }\approx 365*24*3600sT=1 year ≈365∗24∗3600s
v=2π∗1.5∗1011365∗24∗3600=29.89∗103msv= \frac{2\pi*1.5*10^{11}}{365*24*3600}=29.89*10^3\frac{m}{s}v=365∗24∗36002π∗1.5∗1011=29.89∗103sm
a=v2r=29.892∗1061.5∗1011=0.00595ms2a =\frac{v^2}{r}= \frac{29.89^2*10^6}{1.5*10^{11}}=0.00595\frac{m}{s^2}a=rv2=1.5∗101129.892∗106=0.00595s2m
Answer: a=0.00595ms2\text{Answer: }a =0.00595\frac{m}{s^2}Answer: a=0.00595s2m
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