Could we possibly see the shadow offset from Parker Solar probe on Earth?












9














The closer an object to the source of light, the larger the shadow it casts.





Parker Solar probe would fly within 3.7 million miles (6 million kilometers) of the sun's surface. It sounds quite far away; however it's more than eight times closer than any other spacecraft and more than eight times closer than Mercury. Let's say there is a solar eclipse event right above New York City; the city would be underneath the shadow of the body between Earth and Sun. So here's the wonder: with Parker Solar probe is staying that close to the Sun, could we possibly see its shadow offset from it on Earth?










share|improve this question



























    9














    The closer an object to the source of light, the larger the shadow it casts.





    Parker Solar probe would fly within 3.7 million miles (6 million kilometers) of the sun's surface. It sounds quite far away; however it's more than eight times closer than any other spacecraft and more than eight times closer than Mercury. Let's say there is a solar eclipse event right above New York City; the city would be underneath the shadow of the body between Earth and Sun. So here's the wonder: with Parker Solar probe is staying that close to the Sun, could we possibly see its shadow offset from it on Earth?










    share|improve this question

























      9












      9








      9







      The closer an object to the source of light, the larger the shadow it casts.





      Parker Solar probe would fly within 3.7 million miles (6 million kilometers) of the sun's surface. It sounds quite far away; however it's more than eight times closer than any other spacecraft and more than eight times closer than Mercury. Let's say there is a solar eclipse event right above New York City; the city would be underneath the shadow of the body between Earth and Sun. So here's the wonder: with Parker Solar probe is staying that close to the Sun, could we possibly see its shadow offset from it on Earth?










      share|improve this question













      The closer an object to the source of light, the larger the shadow it casts.





      Parker Solar probe would fly within 3.7 million miles (6 million kilometers) of the sun's surface. It sounds quite far away; however it's more than eight times closer than any other spacecraft and more than eight times closer than Mercury. Let's say there is a solar eclipse event right above New York City; the city would be underneath the shadow of the body between Earth and Sun. So here's the wonder: with Parker Solar probe is staying that close to the Sun, could we possibly see its shadow offset from it on Earth?







      the-sun parker-solar-probe






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Dec 3 at 1:14









      Boosted Nub

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      557222






















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          21















          The closer an object to the source of light, the larger the shadow it casts.




          That's true if we're talking about a point source or at least a compact source of light and "shadow" refers to the "umbra" or area of complete shadowing. But it no longer makes sense in this case where seen from Earth the obscurer (spacecraft) is tiny compared to the "obscuree" (Sun).



          enter image description here Source



          In this case we can call the event a transit of the Sun by Parker and can treat it just like a similar transit by Mercury, only smaller.



          At the distance of the Earth, There is no umbra, only an antumbra. Parker's umbra only extends about 250 meters behind the spacecraft's 2.3 meter hexagonal Sun shield.



          In the case of Mercury, let's do the math.



          body      radius (km)     distance (km)   solid angle (sr)   relative to Sun
          Sun 695,000. 150,000,000. 6.7E-05 -
          Mercury 2,440. 92,000,000. 2.1E-09 3.1E-05
          Parker 0.0015 150,000,000. 3.1E-22 4.6E-18


          So while transit of Mercury will dim the Sun everywhere on Earth almost equally by about 31 parts per million (and would be noticed by good quality photometry from a satellite), Parker would only dim the Sun by five quintillionths, which is far lower than normal fluctuations in the Suns brightness.



          At about 195 km/s Parker will transit the Sun's disk in about 7,000 seconds, or 2 hours. Below are examples of how the Sun's brightness fluctuates on this timescale. Certainly a step function of 2E-05 (from Mercury) for hours would be detectable, but one of 3E-12 (from Parker) would be so far in the noise as to be completely undetectable.



          However, if you wanted to measure the solar transit of the ISS instead, using a Raspberry Pi or an Arduino and a photodiode from Earth instead, that's certainly doable because of the shortness of the pulse.



          below x3: An estimate of the solar background irradiance power spectrum Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)




          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)



          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)



          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)







          share|improve this answer



















          • 3




            "There is no umbra" Well, there is; it's just very small on astronomical scales
            – Antzi
            Dec 3 at 2:34






          • 1




            Yes, that's exactly what I meant (would be a nice addition to your already very good answer)
            – Antzi
            Dec 3 at 2:38








          • 1




            Parker has a radius of 1.15 km?
            – Jordi Vermeulen
            Dec 3 at 14:10






          • 1




            @JordiVermeulen ̶y̶e̶s̶,̶ ̶i̶n̶ ̶t̶h̶e̶ ̶v̶a̶c̶u̶u̶m̶ ̶o̶f̶ ̶s̶p̶a̶c̶e̶ ̶i̶t̶ ̶e̶x̶p̶a̶n̶d̶s̶.̶ er, I mean good catch, thanks! How does it look now?
            – uhoh
            Dec 3 at 14:25








          • 2




            @uhoh that looks like a more realistic number. Although it would be cool if it had instrumentation that expanded to a massive size.
            – Jordi Vermeulen
            Dec 3 at 14:32











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          21















          The closer an object to the source of light, the larger the shadow it casts.




          That's true if we're talking about a point source or at least a compact source of light and "shadow" refers to the "umbra" or area of complete shadowing. But it no longer makes sense in this case where seen from Earth the obscurer (spacecraft) is tiny compared to the "obscuree" (Sun).



          enter image description here Source



          In this case we can call the event a transit of the Sun by Parker and can treat it just like a similar transit by Mercury, only smaller.



          At the distance of the Earth, There is no umbra, only an antumbra. Parker's umbra only extends about 250 meters behind the spacecraft's 2.3 meter hexagonal Sun shield.



          In the case of Mercury, let's do the math.



          body      radius (km)     distance (km)   solid angle (sr)   relative to Sun
          Sun 695,000. 150,000,000. 6.7E-05 -
          Mercury 2,440. 92,000,000. 2.1E-09 3.1E-05
          Parker 0.0015 150,000,000. 3.1E-22 4.6E-18


          So while transit of Mercury will dim the Sun everywhere on Earth almost equally by about 31 parts per million (and would be noticed by good quality photometry from a satellite), Parker would only dim the Sun by five quintillionths, which is far lower than normal fluctuations in the Suns brightness.



          At about 195 km/s Parker will transit the Sun's disk in about 7,000 seconds, or 2 hours. Below are examples of how the Sun's brightness fluctuates on this timescale. Certainly a step function of 2E-05 (from Mercury) for hours would be detectable, but one of 3E-12 (from Parker) would be so far in the noise as to be completely undetectable.



          However, if you wanted to measure the solar transit of the ISS instead, using a Raspberry Pi or an Arduino and a photodiode from Earth instead, that's certainly doable because of the shortness of the pulse.



          below x3: An estimate of the solar background irradiance power spectrum Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)




          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)



          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)



          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)







          share|improve this answer



















          • 3




            "There is no umbra" Well, there is; it's just very small on astronomical scales
            – Antzi
            Dec 3 at 2:34






          • 1




            Yes, that's exactly what I meant (would be a nice addition to your already very good answer)
            – Antzi
            Dec 3 at 2:38








          • 1




            Parker has a radius of 1.15 km?
            – Jordi Vermeulen
            Dec 3 at 14:10






          • 1




            @JordiVermeulen ̶y̶e̶s̶,̶ ̶i̶n̶ ̶t̶h̶e̶ ̶v̶a̶c̶u̶u̶m̶ ̶o̶f̶ ̶s̶p̶a̶c̶e̶ ̶i̶t̶ ̶e̶x̶p̶a̶n̶d̶s̶.̶ er, I mean good catch, thanks! How does it look now?
            – uhoh
            Dec 3 at 14:25








          • 2




            @uhoh that looks like a more realistic number. Although it would be cool if it had instrumentation that expanded to a massive size.
            – Jordi Vermeulen
            Dec 3 at 14:32
















          21















          The closer an object to the source of light, the larger the shadow it casts.




          That's true if we're talking about a point source or at least a compact source of light and "shadow" refers to the "umbra" or area of complete shadowing. But it no longer makes sense in this case where seen from Earth the obscurer (spacecraft) is tiny compared to the "obscuree" (Sun).



          enter image description here Source



          In this case we can call the event a transit of the Sun by Parker and can treat it just like a similar transit by Mercury, only smaller.



          At the distance of the Earth, There is no umbra, only an antumbra. Parker's umbra only extends about 250 meters behind the spacecraft's 2.3 meter hexagonal Sun shield.



          In the case of Mercury, let's do the math.



          body      radius (km)     distance (km)   solid angle (sr)   relative to Sun
          Sun 695,000. 150,000,000. 6.7E-05 -
          Mercury 2,440. 92,000,000. 2.1E-09 3.1E-05
          Parker 0.0015 150,000,000. 3.1E-22 4.6E-18


          So while transit of Mercury will dim the Sun everywhere on Earth almost equally by about 31 parts per million (and would be noticed by good quality photometry from a satellite), Parker would only dim the Sun by five quintillionths, which is far lower than normal fluctuations in the Suns brightness.



          At about 195 km/s Parker will transit the Sun's disk in about 7,000 seconds, or 2 hours. Below are examples of how the Sun's brightness fluctuates on this timescale. Certainly a step function of 2E-05 (from Mercury) for hours would be detectable, but one of 3E-12 (from Parker) would be so far in the noise as to be completely undetectable.



          However, if you wanted to measure the solar transit of the ISS instead, using a Raspberry Pi or an Arduino and a photodiode from Earth instead, that's certainly doable because of the shortness of the pulse.



          below x3: An estimate of the solar background irradiance power spectrum Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)




          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)



          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)



          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)







          share|improve this answer



















          • 3




            "There is no umbra" Well, there is; it's just very small on astronomical scales
            – Antzi
            Dec 3 at 2:34






          • 1




            Yes, that's exactly what I meant (would be a nice addition to your already very good answer)
            – Antzi
            Dec 3 at 2:38








          • 1




            Parker has a radius of 1.15 km?
            – Jordi Vermeulen
            Dec 3 at 14:10






          • 1




            @JordiVermeulen ̶y̶e̶s̶,̶ ̶i̶n̶ ̶t̶h̶e̶ ̶v̶a̶c̶u̶u̶m̶ ̶o̶f̶ ̶s̶p̶a̶c̶e̶ ̶i̶t̶ ̶e̶x̶p̶a̶n̶d̶s̶.̶ er, I mean good catch, thanks! How does it look now?
            – uhoh
            Dec 3 at 14:25








          • 2




            @uhoh that looks like a more realistic number. Although it would be cool if it had instrumentation that expanded to a massive size.
            – Jordi Vermeulen
            Dec 3 at 14:32














          21












          21








          21







          The closer an object to the source of light, the larger the shadow it casts.




          That's true if we're talking about a point source or at least a compact source of light and "shadow" refers to the "umbra" or area of complete shadowing. But it no longer makes sense in this case where seen from Earth the obscurer (spacecraft) is tiny compared to the "obscuree" (Sun).



          enter image description here Source



          In this case we can call the event a transit of the Sun by Parker and can treat it just like a similar transit by Mercury, only smaller.



          At the distance of the Earth, There is no umbra, only an antumbra. Parker's umbra only extends about 250 meters behind the spacecraft's 2.3 meter hexagonal Sun shield.



          In the case of Mercury, let's do the math.



          body      radius (km)     distance (km)   solid angle (sr)   relative to Sun
          Sun 695,000. 150,000,000. 6.7E-05 -
          Mercury 2,440. 92,000,000. 2.1E-09 3.1E-05
          Parker 0.0015 150,000,000. 3.1E-22 4.6E-18


          So while transit of Mercury will dim the Sun everywhere on Earth almost equally by about 31 parts per million (and would be noticed by good quality photometry from a satellite), Parker would only dim the Sun by five quintillionths, which is far lower than normal fluctuations in the Suns brightness.



          At about 195 km/s Parker will transit the Sun's disk in about 7,000 seconds, or 2 hours. Below are examples of how the Sun's brightness fluctuates on this timescale. Certainly a step function of 2E-05 (from Mercury) for hours would be detectable, but one of 3E-12 (from Parker) would be so far in the noise as to be completely undetectable.



          However, if you wanted to measure the solar transit of the ISS instead, using a Raspberry Pi or an Arduino and a photodiode from Earth instead, that's certainly doable because of the shortness of the pulse.



          below x3: An estimate of the solar background irradiance power spectrum Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)




          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)



          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)



          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)







          share|improve this answer















          The closer an object to the source of light, the larger the shadow it casts.




          That's true if we're talking about a point source or at least a compact source of light and "shadow" refers to the "umbra" or area of complete shadowing. But it no longer makes sense in this case where seen from Earth the obscurer (spacecraft) is tiny compared to the "obscuree" (Sun).



          enter image description here Source



          In this case we can call the event a transit of the Sun by Parker and can treat it just like a similar transit by Mercury, only smaller.



          At the distance of the Earth, There is no umbra, only an antumbra. Parker's umbra only extends about 250 meters behind the spacecraft's 2.3 meter hexagonal Sun shield.



          In the case of Mercury, let's do the math.



          body      radius (km)     distance (km)   solid angle (sr)   relative to Sun
          Sun 695,000. 150,000,000. 6.7E-05 -
          Mercury 2,440. 92,000,000. 2.1E-09 3.1E-05
          Parker 0.0015 150,000,000. 3.1E-22 4.6E-18


          So while transit of Mercury will dim the Sun everywhere on Earth almost equally by about 31 parts per million (and would be noticed by good quality photometry from a satellite), Parker would only dim the Sun by five quintillionths, which is far lower than normal fluctuations in the Suns brightness.



          At about 195 km/s Parker will transit the Sun's disk in about 7,000 seconds, or 2 hours. Below are examples of how the Sun's brightness fluctuates on this timescale. Certainly a step function of 2E-05 (from Mercury) for hours would be detectable, but one of 3E-12 (from Parker) would be so far in the noise as to be completely undetectable.



          However, if you wanted to measure the solar transit of the ISS instead, using a Raspberry Pi or an Arduino and a photodiode from Earth instead, that's certainly doable because of the shortness of the pulse.



          below x3: An estimate of the solar background irradiance power spectrum Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)




          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)



          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)



          Noise of the Sun Rabello Soares et al. Astron. Astrophys. 318, 970–974 (1997)








          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Dec 3 at 14:22

























          answered Dec 3 at 2:05









          uhoh

          34.3k17117429




          34.3k17117429








          • 3




            "There is no umbra" Well, there is; it's just very small on astronomical scales
            – Antzi
            Dec 3 at 2:34






          • 1




            Yes, that's exactly what I meant (would be a nice addition to your already very good answer)
            – Antzi
            Dec 3 at 2:38








          • 1




            Parker has a radius of 1.15 km?
            – Jordi Vermeulen
            Dec 3 at 14:10






          • 1




            @JordiVermeulen ̶y̶e̶s̶,̶ ̶i̶n̶ ̶t̶h̶e̶ ̶v̶a̶c̶u̶u̶m̶ ̶o̶f̶ ̶s̶p̶a̶c̶e̶ ̶i̶t̶ ̶e̶x̶p̶a̶n̶d̶s̶.̶ er, I mean good catch, thanks! How does it look now?
            – uhoh
            Dec 3 at 14:25








          • 2




            @uhoh that looks like a more realistic number. Although it would be cool if it had instrumentation that expanded to a massive size.
            – Jordi Vermeulen
            Dec 3 at 14:32














          • 3




            "There is no umbra" Well, there is; it's just very small on astronomical scales
            – Antzi
            Dec 3 at 2:34






          • 1




            Yes, that's exactly what I meant (would be a nice addition to your already very good answer)
            – Antzi
            Dec 3 at 2:38








          • 1




            Parker has a radius of 1.15 km?
            – Jordi Vermeulen
            Dec 3 at 14:10






          • 1




            @JordiVermeulen ̶y̶e̶s̶,̶ ̶i̶n̶ ̶t̶h̶e̶ ̶v̶a̶c̶u̶u̶m̶ ̶o̶f̶ ̶s̶p̶a̶c̶e̶ ̶i̶t̶ ̶e̶x̶p̶a̶n̶d̶s̶.̶ er, I mean good catch, thanks! How does it look now?
            – uhoh
            Dec 3 at 14:25








          • 2




            @uhoh that looks like a more realistic number. Although it would be cool if it had instrumentation that expanded to a massive size.
            – Jordi Vermeulen
            Dec 3 at 14:32








          3




          3




          "There is no umbra" Well, there is; it's just very small on astronomical scales
          – Antzi
          Dec 3 at 2:34




          "There is no umbra" Well, there is; it's just very small on astronomical scales
          – Antzi
          Dec 3 at 2:34




          1




          1




          Yes, that's exactly what I meant (would be a nice addition to your already very good answer)
          – Antzi
          Dec 3 at 2:38






          Yes, that's exactly what I meant (would be a nice addition to your already very good answer)
          – Antzi
          Dec 3 at 2:38






          1




          1




          Parker has a radius of 1.15 km?
          – Jordi Vermeulen
          Dec 3 at 14:10




          Parker has a radius of 1.15 km?
          – Jordi Vermeulen
          Dec 3 at 14:10




          1




          1




          @JordiVermeulen ̶y̶e̶s̶,̶ ̶i̶n̶ ̶t̶h̶e̶ ̶v̶a̶c̶u̶u̶m̶ ̶o̶f̶ ̶s̶p̶a̶c̶e̶ ̶i̶t̶ ̶e̶x̶p̶a̶n̶d̶s̶.̶ er, I mean good catch, thanks! How does it look now?
          – uhoh
          Dec 3 at 14:25






          @JordiVermeulen ̶y̶e̶s̶,̶ ̶i̶n̶ ̶t̶h̶e̶ ̶v̶a̶c̶u̶u̶m̶ ̶o̶f̶ ̶s̶p̶a̶c̶e̶ ̶i̶t̶ ̶e̶x̶p̶a̶n̶d̶s̶.̶ er, I mean good catch, thanks! How does it look now?
          – uhoh
          Dec 3 at 14:25






          2




          2




          @uhoh that looks like a more realistic number. Although it would be cool if it had instrumentation that expanded to a massive size.
          – Jordi Vermeulen
          Dec 3 at 14:32




          @uhoh that looks like a more realistic number. Although it would be cool if it had instrumentation that expanded to a massive size.
          – Jordi Vermeulen
          Dec 3 at 14:32


















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