How much work is required to lift a 1400-kg satellite to an altitude of 2⋅106 m above the surface of the Earth? The gravitational force is F=GMm/r2, where M is the mass of the Earth, m is the mass of the satellite, and r is the distance between the satellite and the Earth's center. The radius of the Earth is 6.4⋅106 m, its mass is 6⋅1024 kg, and in these units the gravitational constant, G, is 6.67⋅10−11.

2.08 x 10^10 J.

Step-by-step explanation:

Given:

m = 1400 kg

H = 2 x 10^6 m

R = 6.4 x 10^6 m

M = 6 x 10^24 kg

G = 6.67 x 10^-11

F = GMm/r^2

= GMm/r²

Work needed to lift a mass m from R (the surface) to R+h (height h above the surface) is just the difference of potential energy between those two altitudes. And potential energy is the r-integral of the force:

V(r) = -GMm/r

So you can calculate the lifting work from that.

Work, W = V(R+h) - V(R)

= GMm[1/R - 1/(R+h)]

Inputting values,

W = (6.67 × 10^−11) * (6 × 10^24) * (1400) * [(1/6.4×10^6) - (1/(6.4 × 10^6)+(2 × 10^6))]

= 2.08 x 10^10

W = 2.08 x 10^10 J.

answer

a,b, and d

explanation

a reflection angle is more than

but less than;

an angle that measures

is a reflex angle.(true)

an angle that measures less than

is acute.(true)

provided we are considering angles greater or equal to zero.

an acute angle lies in the range.

an obtuse angle is less than 180° but more than 90°

an angle that measures 300° is obtuse (false)

an angle that measures 65° is an acute angle.(true)

correct answer is option b.

step-by-step explanation:

the true statement about geometric figure is option b that is one line can be drawn through all three points.

if three points are on the same plane, they can be connected by a line.