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ivann1987 [24]
3 years ago
5

A satellite moves in a circular orbit a distance of 1.6×10^5 m above Earth's surface. (radius of Earth is 6.38 x 10^6m and its m

ass is 5.98 x 10^24 kg). Determine the speed of the satellite.
Physics
1 answer:
Rashid [163]3 years ago
5 0

Answer:

The speed of the satellite is 7809.52 m/s        

Explanation:

It is given that,

Radius of Earth, r=6.38\times 10^6\ m

Mass of earth, M=5.98\times 10^{24}\ kg

A satellite moves in a circular orbit a distance of, d=1.6\times 10^5\ m  above Earth's surface.

We need to find the speed of the satellite. It is given by :

v=\sqrt{\dfrac{GM}{R}}

R = r + d

R=(6.38\times 10^6\ m+1.6\times 10^5\ m)=6540000\ m

So, v=\sqrt{\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 5.98\times 10^{24}\ kg}{6540000\ m}}

v = 7809.52 m/s

So, the speed of the satellite is 7809.52 m/s. Hence, this is the required solution.

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Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
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Answer:

160 kg

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Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

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u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

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m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

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