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2 years ago

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On a spacecraft sent to Mars to take pictures, the camera is trigered by radio waves, which, like all electromagnetic waves, tra

vel with the speed of light. The speed of light is 3 × 108 m/s and the distance to Mars is 9.7 × 1010 m. What is the time delay between sending the signal from Earth and receiving the signal on Mars?

1 answer:

alex41 [277]2 years ago

8 0

Kinematic equations of linear motion tell us that speed is equivalent to distance traveled over a period of time, that is

$v = \frac{x}{t}$

Here,

x = Distance

t = Time

Rearranging to find the time we have that,

$t = \frac{x}{v}$

The speed is equal to the speed of light ( $3*10^8m/s$ ) and the distance (x) is ( $9.7*10^{10} m$ )

$t= \frac{9.7*10^{10} }{3*10^8}$

$t = 323.33s$

The time delay between sending the signal from Earth and receiving the signal on Mars is 323.33s

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An astronaut weighs 8.00 × 102 newtons on the sur- face of Earth. What is the weight of the astronaut 6.37 × 106 meters above th

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We know the weight of the astronout on the surface, with this we can find his mass. Letting $w_{s}$ be the weight on the surface:

$w_{s}=mg$ ,

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since we now that $g=9.8m/s^{2}$ we get that the mass is

$m=81.6kg$ .

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$F=G\frac{Mm}{r^{2}}$ ,

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$F=ma$ ,

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and this get us to

$mg=G\frac{Mm}{r^{2}}$ , where $mg$ is his new weight.

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The total distance between the astronaut and the earth is

$r=(6.37*10^{6}+6.37*10^{6})=2(6.37*10^{6})=12.74*10^{6}$ meters.

Now we can compute his weigh:

$mg=G\frac{Mm}{r^{2}}$ ,

$mg=(6.674*10^{-11})\frac{(5.972*10^{24})(81.6)}{(12.74*10^{6})^{2}}$ ,

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