Question

1.The space shuttle is in a 300km -high circular orbit. It needs to reach a 640km -high circular orbit to catch the Hubble Space Telescope for repairs. The shuttle's mass is 8.00

Answer 1

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1.

The space shuttle is in a 950 km high circular orbit. It needs to reach a 610 km high circular orbit to catch the hubble space telescope for repairs. The shuttle mass is 75,000 kg. How much energy is required to boost it to the new orbit?

The formula for orbital energy is E=-(G*m*M)/(2R), where
G = gravitational constant =6,67 *10^-11 m^3/(kg*s^2)
m - mass of the shuttle
M- mass of the Earth
R- distance from the centre of the Earth to the shuttle (I don't know if orbit in your case is given from the centre or from the surface, so I will calculate as from the centre. if not just add the radius of the Earth to the number R)

So at the beginning
m=75000kg
M=5.97 * 10^24 kg (find inliteraturee)
R=950*10^3 m

E1=-(6,67 *10^-11 *75000*5.97 * 10^24 )/(2*950*10^3)=-1.57*10^13 J

In the end
R=610000 m
E2=-(6,67*10^-11 *75000*5.97 * 10^24 )/(2*610*10^3)=-2.45*10^13 J

So W=E2-E1=(-2.45*10^13) - (-1.57*10^13) =-0.88*10^13 J
You can loose the - sign if you want, that is just in theory assumed, that the object has negative energy.

2.

Maybe the radius is from the (earth + moon) centre of mass (barycentre) ?

G = 6.67384 e-11 (a constant)
M = Earth mass + moon mass = 6.045 e24 kg
R = distance from barycentre to far side surface of the moon = 3.815 e8 metres

For theoretical escape velocity (Ve) ( projectile mass = 0 kg ) :

Ve = sqrt ( ( 2 * G * M ) / R ) (see also note below)

Ve = 1,454.42 m/s

Notes :
If you give the projectile a mass, the Ve will increase, though the mass needs to be substantial to be noticeable, and this would be a true two body problem and more sophisticated.
(i ran this with a negligable mass value for m in a two body program, and the answer is the same as above, though theres a difference if you use enough decimal places)

The distance from the earths centre to the system barycentre = 4.67197 e6 metres.

The equation G M m / r =