Why does Starman/Roadster have radial acceleration?
$begingroup$
Discussion in chat pointed out that the SpaceX Roadster or Starman has a Marsden A1 coefficient.
From solution 10 in Horizons:
EPOCH= 2458164.5
EC= .2585469914787243 QR= .9860596231806226 TP= 2458153.620483722645
OM= 317.3549094214575 W = 177.3203028023227 IN= 1.088451292866039
A1= 2.960683526738534E-9 R0= 1. ALN= 1. NM= 2. NK= 0.
SRC= -2.057839421666802E-7...
I wrote about the Marsden parameters in this question a while ago and have linked several sources there. You have to use that in a certain power-law equation and use units of days and AU, so A1 is 2.96E-09 AU/day^2.
Question: Why does Starman/Roadster have radial acceleration? What is causing this? Can the size of the acceleration be understood quantitatively?
Aliens are trying, very slowly, to steal it?
Air leaking out of the tires?
orbital-mechanics spacex tesla-roadster
edited 8 hours ago
PearsonArtPhoto♦
82.7k16235450
asked 13 hours ago
uhohuhoh
37.6k18136480
$endgroup$
add a comment |
$begingroup$
Discussion in chat pointed out that the SpaceX Roadster or Starman has a Marsden A1 coefficient.
From solution 10 in Horizons:
EPOCH= 2458164.5
EC= .2585469914787243 QR= .9860596231806226 TP= 2458153.620483722645
OM= 317.3549094214575 W = 177.3203028023227 IN= 1.088451292866039
A1= 2.960683526738534E-9 R0= 1. ALN= 1. NM= 2. NK= 0.
SRC= -2.057839421666802E-7...
I wrote about the Marsden parameters in this question a while ago and have linked several sources there. You have to use that in a certain power-law equation and use units of days and AU, so A1 is 2.96E-09 AU/day^2.
Question: Why does Starman/Roadster have radial acceleration? What is causing this? Can the size of the acceleration be understood quantitatively?
Aliens are trying, very slowly, to steal it?
Air leaking out of the tires?
orbital-mechanics spacex tesla-roadster
edited 8 hours ago
PearsonArtPhoto♦
82.7k16235450
asked 13 hours ago
uhohuhoh
37.6k18136480
$endgroup$
add a comment |
$begingroup$
Discussion in chat pointed out that the SpaceX Roadster or Starman has a Marsden A1 coefficient.
From solution 10 in Horizons:
EPOCH= 2458164.5
EC= .2585469914787243 QR= .9860596231806226 TP= 2458153.620483722645
OM= 317.3549094214575 W = 177.3203028023227 IN= 1.088451292866039
A1= 2.960683526738534E-9 R0= 1. ALN= 1. NM= 2. NK= 0.
SRC= -2.057839421666802E-7...
I wrote about the Marsden parameters in this question a while ago and have linked several sources there. You have to use that in a certain power-law equation and use units of days and AU, so A1 is 2.96E-09 AU/day^2.
Question: Why does Starman/Roadster have radial acceleration? What is causing this? Can the size of the acceleration be understood quantitatively?
Aliens are trying, very slowly, to steal it?
Air leaking out of the tires?
orbital-mechanics spacex tesla-roadster
edited 8 hours ago
PearsonArtPhoto♦
82.7k16235450
asked 13 hours ago
uhohuhoh
37.6k18136480
$endgroup$
Discussion in chat pointed out that the SpaceX Roadster or Starman has a Marsden A1 coefficient.
From solution 10 in Horizons:
EPOCH= 2458164.5
EC= .2585469914787243 QR= .9860596231806226 TP= 2458153.620483722645
OM= 317.3549094214575 W = 177.3203028023227 IN= 1.088451292866039
A1= 2.960683526738534E-9 R0= 1. ALN= 1. NM= 2. NK= 0.
SRC= -2.057839421666802E-7...
I wrote about the Marsden parameters in this question a while ago and have linked several sources there. You have to use that in a certain power-law equation and use units of days and AU, so A1 is 2.96E-09 AU/day^2.
Question: Why does Starman/Roadster have radial acceleration? What is causing this? Can the size of the acceleration be understood quantitatively?
Aliens are trying, very slowly, to steal it?
Air leaking out of the tires?
orbital-mechanics spacex tesla-roadster
orbital-mechanics spacex tesla-roadster
edited 8 hours ago
PearsonArtPhoto♦
82.7k16235450
asked 13 hours ago
uhohuhoh
37.6k18136480
edited 8 hours ago
PearsonArtPhoto♦
82.7k16235450
asked 13 hours ago
uhohuhoh
37.6k18136480
edited 8 hours ago
PearsonArtPhoto♦
82.7k16235450
edited 8 hours ago
PearsonArtPhoto♦
82.7k16235450
edited 8 hours ago
PearsonArtPhoto♦
82.7k16235450
82.7k16235450
asked 13 hours ago
uhohuhoh
37.6k18136480
asked 13 hours ago
uhohuhoh
37.6k18136480
asked 13 hours ago
uhohuhoh
37.6k18136480
37.6k18136480
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Sunlight pressure. The acceleration is 9.08 μN / m2 (Assuming perfect reflectivity). The size is about 3.66* 12.6 = 46 m2. That gives a thrust of about 414 uN. The mass is about 1300 kg. Thus the acceleration from sunlight max is about 3.2 e-7 m/s2. Of course, there are a lot of assumptions in that, the mass is probably higher, it won't be perfect reflectivity, and a good part of that time not all of it will be pointed in the correct direction, still, this is a fair estimate. The acceleration will also vary as one gets further from the Sun.
The units for A1 from Horizons are actually AU/day2, and are subject to a very complex formula. That changes to 5.93321266 × 10-8 m/s2. It is even more complex than that, however, the actual acceleration depends on the distance from the Sun. Lucky for us, the numbers I calculated assume 1 AU, and at that point they are nearly identical. So that is a reasonable approximation, and this value is perfectly consistent with a low reflective pointing mostly away from the Sun Starman.
edited 8 hours ago
Community♦
1
answered 13 hours ago
PearsonArtPhoto♦PearsonArtPhoto
82.7k16235450
$endgroup$
$begingroup$
46 m^2, not m^s.
$endgroup$
– Skyler
10 hours ago
$begingroup$
Doesn't the solar wind also provide some outward pressure?
$endgroup$
– Loren Pechtel
9 mins ago
add a comment |
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1 Answer
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oldest
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
Sunlight pressure. The acceleration is 9.08 μN / m2 (Assuming perfect reflectivity). The size is about 3.66* 12.6 = 46 m2. That gives a thrust of about 414 uN. The mass is about 1300 kg. Thus the acceleration from sunlight max is about 3.2 e-7 m/s2. Of course, there are a lot of assumptions in that, the mass is probably higher, it won't be perfect reflectivity, and a good part of that time not all of it will be pointed in the correct direction, still, this is a fair estimate. The acceleration will also vary as one gets further from the Sun.
The units for A1 from Horizons are actually AU/day2, and are subject to a very complex formula. That changes to 5.93321266 × 10-8 m/s2. It is even more complex than that, however, the actual acceleration depends on the distance from the Sun. Lucky for us, the numbers I calculated assume 1 AU, and at that point they are nearly identical. So that is a reasonable approximation, and this value is perfectly consistent with a low reflective pointing mostly away from the Sun Starman.
edited 8 hours ago
Community♦
1
answered 13 hours ago
PearsonArtPhoto♦PearsonArtPhoto
82.7k16235450
$endgroup$
$begingroup$
46 m^2, not m^s.
$endgroup$
– Skyler
10 hours ago
$begingroup$
Doesn't the solar wind also provide some outward pressure?
$endgroup$
– Loren Pechtel
9 mins ago
add a comment |
$begingroup$
Sunlight pressure. The acceleration is 9.08 μN / m2 (Assuming perfect reflectivity). The size is about 3.66* 12.6 = 46 m2. That gives a thrust of about 414 uN. The mass is about 1300 kg. Thus the acceleration from sunlight max is about 3.2 e-7 m/s2. Of course, there are a lot of assumptions in that, the mass is probably higher, it won't be perfect reflectivity, and a good part of that time not all of it will be pointed in the correct direction, still, this is a fair estimate. The acceleration will also vary as one gets further from the Sun.
The units for A1 from Horizons are actually AU/day2, and are subject to a very complex formula. That changes to 5.93321266 × 10-8 m/s2. It is even more complex than that, however, the actual acceleration depends on the distance from the Sun. Lucky for us, the numbers I calculated assume 1 AU, and at that point they are nearly identical. So that is a reasonable approximation, and this value is perfectly consistent with a low reflective pointing mostly away from the Sun Starman.
edited 8 hours ago
Community♦
1
answered 13 hours ago
PearsonArtPhoto♦PearsonArtPhoto
82.7k16235450
$endgroup$
$begingroup$
46 m^2, not m^s.
$endgroup$
– Skyler
10 hours ago
$begingroup$
Doesn't the solar wind also provide some outward pressure?
$endgroup$
– Loren Pechtel
9 mins ago
add a comment |
$begingroup$
Sunlight pressure. The acceleration is 9.08 μN / m2 (Assuming perfect reflectivity). The size is about 3.66* 12.6 = 46 m2. That gives a thrust of about 414 uN. The mass is about 1300 kg. Thus the acceleration from sunlight max is about 3.2 e-7 m/s2. Of course, there are a lot of assumptions in that, the mass is probably higher, it won't be perfect reflectivity, and a good part of that time not all of it will be pointed in the correct direction, still, this is a fair estimate. The acceleration will also vary as one gets further from the Sun.
The units for A1 from Horizons are actually AU/day2, and are subject to a very complex formula. That changes to 5.93321266 × 10-8 m/s2. It is even more complex than that, however, the actual acceleration depends on the distance from the Sun. Lucky for us, the numbers I calculated assume 1 AU, and at that point they are nearly identical. So that is a reasonable approximation, and this value is perfectly consistent with a low reflective pointing mostly away from the Sun Starman.
edited 8 hours ago
Community♦
1
answered 13 hours ago
PearsonArtPhoto♦PearsonArtPhoto
82.7k16235450
$endgroup$
Sunlight pressure. The acceleration is 9.08 μN / m2 (Assuming perfect reflectivity). The size is about 3.66* 12.6 = 46 m2. That gives a thrust of about 414 uN. The mass is about 1300 kg. Thus the acceleration from sunlight max is about 3.2 e-7 m/s2. Of course, there are a lot of assumptions in that, the mass is probably higher, it won't be perfect reflectivity, and a good part of that time not all of it will be pointed in the correct direction, still, this is a fair estimate. The acceleration will also vary as one gets further from the Sun.
The units for A1 from Horizons are actually AU/day2, and are subject to a very complex formula. That changes to 5.93321266 × 10-8 m/s2. It is even more complex than that, however, the actual acceleration depends on the distance from the Sun. Lucky for us, the numbers I calculated assume 1 AU, and at that point they are nearly identical. So that is a reasonable approximation, and this value is perfectly consistent with a low reflective pointing mostly away from the Sun Starman.
edited 8 hours ago
Community♦
1
answered 13 hours ago
PearsonArtPhoto♦PearsonArtPhoto
82.7k16235450
edited 8 hours ago
Community♦
1
edited 8 hours ago
Community♦
1
edited 8 hours ago
Community♦
1
1
answered 13 hours ago
PearsonArtPhoto♦PearsonArtPhoto
82.7k16235450
answered 13 hours ago
PearsonArtPhoto♦PearsonArtPhoto
82.7k16235450
answered 13 hours ago
PearsonArtPhoto♦PearsonArtPhoto
82.7k16235450
82.7k16235450
$begingroup$
46 m^2, not m^s.
$endgroup$
– Skyler
10 hours ago
$begingroup$
Doesn't the solar wind also provide some outward pressure?
$endgroup$
– Loren Pechtel
9 mins ago
add a comment |
$begingroup$
46 m^2, not m^s.
$endgroup$
– Skyler
10 hours ago
$begingroup$
Doesn't the solar wind also provide some outward pressure?
$endgroup$
– Loren Pechtel
9 mins ago
$begingroup$
46 m^2, not m^s.
$endgroup$
– Skyler
10 hours ago
$begingroup$
46 m^2, not m^s.
$endgroup$
– Skyler
10 hours ago
$begingroup$
Doesn't the solar wind also provide some outward pressure?
$endgroup$
– Loren Pechtel
9 mins ago
$begingroup$
Doesn't the solar wind also provide some outward pressure?
$endgroup$
– Loren Pechtel
9 mins ago
add a comment |
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